Dataset Viewer
problem_id
stringlengths 12
45
| cp_id
stringlengths 2
4
| name
stringlengths 3
34
| description
stringlengths 940
4.03k
| problem_level
stringclasses 4
values | problem_link
stringlengths 56
58
| contest_link
stringclasses 45
values | description_no_samples
stringlengths 713
3.03k
| samples
listlengths 1
3
| num_samples
int64 1
3
| solution_python3
stringlengths 106
4.14k
| solution_english
stringlengths 154
7.91k
| tests
listlengths 9
23
| num_tests
int64 9
23
| runtime_limit_sec
int64 2
15
| memory_limit_mb
int64 256
512
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1333_platinum_good_bitstrings
|
1333
|
Good Bitstrings
|
For any two positive integers $a$ and $b$, define the function
$\texttt{gen_string}(a,b)$ by the following Python code:
def gen_string(a: int, b: int):
res = ""
ia, ib = 0, 0
while ia + ib < a + b:
if ia * b <= ib * a:
res += '0'
ia += 1
else:
res += '1'
ib += 1
return res
Equivalent C++ code:
string gen_string(int64_t a, int64_t b) {
string res;
int ia = 0, ib = 0;
while (ia + ib < a + b) {
if ((__int128)ia * b <= (__int128)ib * a) {
res += '0';
ia++;
} else {
res += '1';
ib++;
}
}
return res;
}
$ia$ will equal $a$ and $ib$ will equal $b$ when the loop terminates, so this
function returns a bitstring of length $a+b$ with exactly $a$ zeroes and $b$
ones. For example, $\texttt{gen_string}(4,10)=01110110111011$.
Call a bitstring $s$ $\textbf{good}$ if there exist positive integers $x$ and
$y$ such that $s=\texttt{gen_string}(x,y)$. Given two positive integers $A$ and
$B$ ($1\le A,B\le 10^{18}$), your job is to compute the number of good prefixes
of $\texttt{gen_string}(A,B)$. For example, there are $6$ good prefixes of
$\texttt{gen_string}(4,10)$:
x = 1 | y = 1 | gen_string(x, y) = 01
x = 1 | y = 2 | gen_string(x, y) = 011
x = 1 | y = 3 | gen_string(x, y) = 0111
x = 2 | y = 5 | gen_string(x, y) = 0111011
x = 3 | y = 7 | gen_string(x, y) = 0111011011
x = 4 | y = 10 | gen_string(x, y) = 01110110111011
INPUT FORMAT (input arrives from the terminal / stdin):
The first line contains $T$ ($1\le T\le 10$), the number of independent test
cases.
Each of the next $T$ lines contains two integers $A$ and $B$.
OUTPUT FORMAT (print output to the terminal / stdout):
The answer for each test case on a new line.
SAMPLE INPUT:
6
1 1
3 5
4 7
8 20
4 10
27 21
SAMPLE OUTPUT:
1
5
7
10
6
13
SCORING:
Input 2: $A,B\le 100$Input 3: $A,B\le 1000$Inputs 4-7: $A,B\le 10^6$Inputs 8-13: All answers are at most $10^5$.Inputs 14-21: No additional constraints.
Problem credits: Benjamin Qi
|
platinum
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1333
|
http://www.usaco.org/index.php?page=open23results
|
For any two positive integers $a$ and $b$, define the function
$\texttt{gen_string}(a,b)$ by the following Python code:
def gen_string(a: int, b: int):
res = ""
ia, ib = 0, 0
while ia + ib < a + b:
if ia * b <= ib * a:
res += '0'
ia += 1
else:
res += '1'
ib += 1
return res
Equivalent C++ code:
string gen_string(int64_t a, int64_t b) {
string res;
int ia = 0, ib = 0;
while (ia + ib < a + b) {
if ((__int128)ia * b <= (__int128)ib * a) {
res += '0';
ia++;
} else {
res += '1';
ib++;
}
}
return res;
}
$ia$ will equal $a$ and $ib$ will equal $b$ when the loop terminates, so this
function returns a bitstring of length $a+b$ with exactly $a$ zeroes and $b$
ones. For example, $\texttt{gen_string}(4,10)=01110110111011$.
Call a bitstring $s$ $\textbf{good}$ if there exist positive integers $x$ and
$y$ such that $s=\texttt{gen_string}(x,y)$. Given two positive integers $A$ and
$B$ ($1\le A,B\le 10^{18}$), your job is to compute the number of good prefixes
of $\texttt{gen_string}(A,B)$. For example, there are $6$ good prefixes of
$\texttt{gen_string}(4,10)$:
x = 1 | y = 1 | gen_string(x, y) = 01
x = 1 | y = 2 | gen_string(x, y) = 011
x = 1 | y = 3 | gen_string(x, y) = 0111
x = 2 | y = 5 | gen_string(x, y) = 0111011
x = 3 | y = 7 | gen_string(x, y) = 0111011011
x = 4 | y = 10 | gen_string(x, y) = 01110110111011
INPUT FORMAT (input arrives from the terminal / stdin):
The first line contains $T$ ($1\le T\le 10$), the number of independent test
cases.
Each of the next $T$ lines contains two integers $A$ and $B$.
OUTPUT FORMAT (print output to the terminal / stdout):
The answer for each test case on a new line.
|
[
{
"input": "6\n1 1\n3 5\n4 7\n8 20\n4 10\n27 21",
"output": "1\n5\n7\n10\n6\n13",
"explanation": "",
"input_explanation": "",
"output_explanation": ""
}
] | 1 |
def solve(a, b):
crs_below, crs_above = b, a # cross(f_1, f'), cross(f', f_2)
ans = 0
on_y_axis = True
while True:
if crs_below > crs_above: # f_1 = f_1 + f_2
mul = (crs_below - 1) // crs_above
ans += mul * (1 + (not on_y_axis))
crs_below -= mul * crs_above
else: # f_2 = f_1 + f_2
mul = crs_above // crs_below
ans += mul
crs_above -= mul * crs_below
on_y_axis = False
if crs_above == 0:
break
return ans + 2 * (crs_below - 1)
T = int(input())
for _ in range(T):
a, b = map(int, input().split())
print(solve(a, b))
|
(Analysis by Benjamin Qi, Reviewed by Richard Qi)
Note: The model solutions for all subtasks are very short, though
understanding why they work is not easy.
Suppose we are trying to compute the answer for $A=a$ and $B=b$.
Subtask 2: $O((a+b)^2)$
Let $s=\texttt{gen_string}(a,b)$. For each prefix of $s$, count the number of 0s
and 1s (let these be $c$ and $d$, respectively), and then check whether
$\texttt{gen_string}(c, d)$ is a prefix of $s$.
Subtask 3: $O(a+b)$
The first step is to treat each bit string as a path on the upper-right quadrant
of a 2D grid (all lattice points $(x,y)$ satisfying $x\ge 0$ and $y\ge 0$).
Henceforth, we use "point" as shorthand for "lattice point." Starting at the
point $(0,0)$, we repeatedly move right if we are on or above the line
$y=b/a\cdot x$, and up otherwise, until we reach the point $(a,b)$. This is
equivalent to the function provided in the problem statement because the
condition $ia * b \le ib * a$ compares the slope of the line from the origin to
$(ia, ib)$ with the slope of the line from the origin to $(a, b)$.
Definition: Say that a point $(x,y)$ with $0<x\le a$ and $0<y\le b$ is
good if $\texttt{gen_string}(x,y)$ is a prefix of
$\texttt{gen_string}(a,b)$.
Our goal is to count the number of good points.
Condition: A point $(x,y)$ is good if and only if every point
$(x_p,y_p)$ in the upper-right quadrant satisfying $0\le x_p<x$ or $0\le y_p<y$
is on the same side of the lines through the origin with slopes $y/x$ and
$b/a$. Specifically, $(x_p,y_p)$ is either above or on both lines, or below both
lines.
Proof: Consider pairing the steps of $\texttt{gen_string}(x,y)$ and the
first $x+y$ steps of $\texttt{gen_string}(a,b)$. The given condition is
sufficient to ensure that every pair of steps moves in the same direction. For
the other direction, observe if both steps in a pair move from $(c,d)$ to
$(c,d+1)$, then every point $(x_p,y_p)$ with $y_p=d$ and $x_p\ge c$ is below
the line, while if both steps in a pair move from $(c,d)$ to $(c+1,d)$, then
every point $(x_p,y_p)$ with $x_p=c$ and $y_p\ge d$ is above on or on the line.
Necessity follows from taking the union of these statements for all $(c,d)$
along the path from $(0,0)$ to $(x,y)$. $\blacksquare$
This condition isn't immediately useful because it tells us to check an infinite
number of points to see whether $(x,y)$ is good. The following corollary tells
us that actually, we only need to check a finite number of points.
Corollary: A point $(x,y)$ is good if and only if every point
$(x_p,y_p)\neq (x,y)$ in the upper-right quadrant satisfying $0\le x_p\le x$
and $0\le y_p\le y$ is on the same side of the lines through the origin
with slopes $y/x$ and $b/a$. We call this set of points the bounding
rectangle associated with $(x,y)$.
We use this corollary to separately count good points below the ray from the
origin through $(a,b)$ and good points on or above this ray.
A point $(x,y)$ below the ray through $(a,b)$ is good if and only if
$(x-1,y)$ is not strictly below the ray through $(a,b)$, and there exist no
points with smaller y-coordinate that lie on or above the ray through $(x,y)$
and below the ray through $(a,b)$.A point $(x,y)$ on or above the ray through $(a,b)$ is good if and only if
$(x,y-1)$ is not on or above the ray through $(a,b)$, and there exist no points
with smaller x-coordinate that lie on or above the ray through $(a,b)$ and below
the ray through
$(x,y)$.
To count good points of the first form, note that for every y-coordinate, there
is at most one point with that y-coordinate that can potentially be good.
Iterate over all y-coordinates from $1$ to $b-1$ in increasing order and find
the leftmost point with that y-coordinate that is below the ray through $(a,b)$.
If the last good point we found is on or above the ray through the current
point, then the current point cannot be good. Otherwise, the current point
produces the greatest slope through the origin out of all points below the ray
with y-coordinate less than or equal to the current y-coordinate, so it is good.
Counting good points of the second form can be done similarly.
Implementation Note: We can check whether a point lies above a ray
without division using the
cross
product operation.
Equivalent Python code (though this isn't fast enough to pass the subtask):
Subtask 4: $O(answer)$
Suppose that we want to generate the good pairs in increasing order of size. If
we look at the good pairs from the sample explanation, we can see that every one
of them is the sum of two previous good pairs (if we treat $(1,0)$ and $(0,1)$
as good). For example, $(1, 2)+(1, 3)=(2, 5)$ and $(1, 2)+(2, 5)=(3,7)$. Why is
this happening?
To show this, let's make some additional observations. Let $f_1=(a_1,b_1)$ and
$f_2=(a_2,b_2)$ be any two points in the upper-right quadrant satisfying
$cross(f_1,f_2)=a_1b_2-a_2b_1=1$. Note that this implies $b_1/a_1<b_2/a_2$, as
we can divide both sides by $a_1a_2$ (here we assume $1/0=\infty$).
Furthermore, by Pick's
theorem, no point lies strictly within the triangle with $(0,0)$, $f_1$, and
$f_2$ as vertices (this triangle has $A=cross(f_1,f_2)/2=1/2, i=0, b=3$).
Fact: A point $f'=(a',b')$ lies on or between the rays from the origin
through $f_1$ and $f_2$ ($b_1/a_1\le b'/a'\le b_2/a_2$) if and only if $f'$ can
be written as a non-negative integer multiple of $f_1$ plus a non-negative
integer multiple of $f_2$.
Proof: The "if" direction is obvious. For the "only if" direction, we may
write $f'=cross(f_1, f')\cdot f_2 + cross(f', f_2)\cdot f_1$. Equality holds
because
$$\begin{align*}
&cross(f_1,cross(f_1, f')\cdot f_2 + cross(f', f_2)\cdot f_1)\\
&=cross(f_1, cross(f_1, f')\cdot f_2)+cross(f_1, cross(f', f_2)\cdot f_1)\\
&=cross(f_1, f')\cdot cross(f_1,f_2)+cross(f',f_2)\cdot cross(f_1,f_1)\\
&=cross(f_1,f'),
\end{align*}$$
where we have used that the cross product is bilinear (linear in each of its
arguments), and similarly,
$cross(f',f_2)=cross(cross(f_1, f')\cdot f_2 + cross(f', f_2)\cdot f_1, f_2).$
Both $cross(f_1, f')$ and $cross(f',f_2)$ are non-negative integers because $f'$
lies on or between the rays. Alternatively, we can loop the following steps
until termination occurs with $f_2=(a'/\gcd(a',b'), b'/\gcd(a',b'))$.
If $f'$ is a multiple of either $f_1$ or $f_2$, we're done.Otherwise, let $f_3=f_1+f_2$. Note that $cross(f_
|
[
{
"input": "6\n1 1\n3 5\n4 7\n8 20\n4 10\n27 21\n",
"output": "1\n5\n7\n10\n6\n13\n"
},
{
"input": "10\n85 13\n71 66\n28 66\n29 71\n96 63\n38 31\n17 68\n36 58\n36 4\n63 33\n",
"output": "20\n32\n12\n15\n27\n16\n36\n12\n15\n17\n"
},
{
"input": "10\n906 527\n608 244\n263 174\n839 281\n153 492\n987 849\n701 390\n195 518\n410 59\n48 971\n",
"output": "23\n42\n50\n80\n19\n31\n28\n32\n30\n34\n"
},
{
"input": "10\n147825 752503\n960741 512108\n83790 812992\n328545 187606\n182751 847169\n818963 814953\n706940 588194\n981994 867047\n248103 887086\n501412 307158\n",
"output": "88\n68\n62\n156\n76\n446\n98\n58\n68\n80\n"
},
{
"input": "10\n681404 333428\n181650 647050\n229028 687782\n657455 275098\n58398 718183\n365468 24732\n503720 479173\n556360 20233\n694125 203877\n617154 619370\n",
"output": "95\n141\n363\n54\n148\n60\n88\n232\n85\n564\n"
},
{
"input": "10\n404235 140228\n733083 982834\n604701 282642\n318592 567367\n846008 887551\n189157 486078\n378169 480062\n408938 400339\n571460 63131\n310407 774407\n",
"output": "126\n76\n73\n49\n88\n58\n60\n129\n178\n127\n"
},
{
"input": "10\n405251 463150\n549854 432108\n808582 443624\n49509 52183\n361422 938598\n776469 624844\n644904 51183\n797412 713474\n254877 321065\n407957 46299\n",
"output": "1403\n108\n66\n68\n84\n68\n1172\n21006\n46\n68\n"
},
{
"input": "10\n663353776591300896 752020665300205866\n943526717545269366 300114424946521359\n230713640398918611 187010130850175053\n118239107794585724 200822570858719651\n723715408292350264 396254911812807997\n704994781514794709 819815153424097044\n296759919488427246 945994708325740034\n927030050569777885 705328510163213131\n451337514521889026 808330792404250933\n789874160488838513 814559672581824706\n",
"output": "576\n314\n1682\n287\n480\n899\n322\n1005\n664\n627\n"
},
{
"input": "10\n263959270418486736 511944810471618212\n719955631781742290 884235488590662642\n911958183634412763 115875852572241461\n882870749898577341 896059443335458468\n497522514232414628 118311430658869194\n953631769711592758 442130823895113645\n516877924612139121 25792863217482507\n670853595357487791 22108287664915026\n610788629622193538 221128154162799165\n410465679038670103 498400723790707670\n",
"output": "503\n186\n463\n679\n327\n331\n229\n269\n269\n1664\n"
},
{
"input": "10\n213394634718308650 567314195893526830\n118718333481546670 984222666619725683\n521169680133050475 379632545405725706\n529433210710147101 666030768028277123\n248727345489185865 923284915041497532\n555027965912713377 803887041840034130\n349121089908475057 822348686848065296\n607629682756169018 642020332228675401\n946903533858245439 44225285843701573\n981228638057593 912509743147346586\n",
"output": "276\n244\n207\n472\n1104\n333\n1074\n288\n590\n1498\n"
},
{
"input": "10\n975925299289124420 60303066417239296\n766474983477708842 896147119677109501\n37292938145223081 40635895974758701\n202932421403534375 409995075828727680\n269767950521665045 612979965315330039\n648872061480630731 203238724068557061\n515872085687282128 932457502235646100\n752197651492851645 591490401816131517\n587106752153443303 7448235807550636\n666789996025594432 670835062832869250\n",
"output": "575\n184\n543\n488\n2313\n283\n302\n974\n264\n322\n"
},
{
"input": "10\n834598416537042887 606025100334242999\n320428844366792646 151368673495529221\n842633031722285124 117239895104072930\n625021458914788036 949669118025122161\n486067279580042617 962611646795744707\n819755947652456438 427518443947948205\n34030594912010556 588251327521452796\n272271618017091404 383321086957162463\n694958646991121366 855790750751112435\n216210753933316896 129209902788953454\n",
"output": "479\n1770\n279\n264\n257\n236\n315\n242\n267\n2859\n"
},
{
"input": "10\n397193281271954503 61672130855956143\n314492482187282527 654066356498811443\n860745630666775130 569333826127128192\n476297769468002128 352294660078528480\n517761936701389633 534752001708683063\n674202688566846814 64972659840190297\n585800934763526200 297942354678997984\n106591030796636628 311626539461044772\n579688750742413520 199518632813615745\n534939827260343216 51723938853748274\n",
"output": "248\n242\n555\n292\n244\n272\n534\n228\n356\n398\n"
},
{
"input": "10\n208 565667425097819020\n55 157321119038912352\n316907676466271933 526\n886500984850879616 22\n299149363183826900 250\n987113150500653689 61\n159 670025876873085627\n937166187092977629 764\n874753340780245712 360\n214921484777387769 800038093746969281\n",
"output": "2719554928354919\n2860383982525694\n602486076932104\n40295499311403632\n1196597452735411\n16182182795092699\n4213999225616910\n1226657312948958\n2429870391056278\n885\n"
},
{
"input": "10\n166972379813397449 29\n569066341265273971 539\n506 968039217663997486\n352182654566540336 969\n121125523365707181 848\n517719135958060282 793\n738 784126291767169684\n298484835440867551 651\n603 365907938824348123\n910247730461828965 805051032603503065\n",
"output": "5757668269427513\n1055781709212057\n1913120983525707\n363449591915970\n142836702082232\n652861457702537\n1062501750362040\n458502051368495\n606812502196291\n1467\n"
},
{
"input": "10\n553486341350597216 428\n126 131864838709694768\n488211818967176514 885\n830 514390130212884291\n548964274098890426 286\n621370290006585458 204\n257458383843435158 856\n758 307692053217471135\n439 434604040461145810\n943589164042917917 667787609709087533\n",
"output": "1293192386333213\n1046546338965896\n551651772844308\n619747144834823\n1919455503842295\n3045932794149964\n300769140004081\n405926191579807\n989986424740685\n230\n"
},
{
"input": "10\n320654112354370136 755\n681 573231478158788514\n375 855899714394872507\n424 703958346499687295\n143 455080993131748359\n696 741114359393839020\n332216328717809233 828\n865773491535017948 263\n480 875075196649813697\n856882629296459076 668553697725070634\n",
"output": "424707433582256\n841749600820568\n2282399238386346\n1660279119103072\n3182384567354903\n1064819481887741\n401227450142320\n3291914416483008\n1823073326353802\n280\n"
},
{
"input": "10\n450591848507395078 766\n970498285745258150 887\n867038654838117493 714\n443 698101902806352737\n387 449758430496187790\n928952692836028738 882\n378854113906172456 822\n74 220142603208608569\n442341090408329812 239\n44892686468418987 155998982976513680\n",
"output": "588240011106301\n1094135609633944\n1214339852714501\n1575850796402627\n1162166487070309\n1053234345619137\n460893082610948\n2974900043359591\n1850799541457469\n282\n"
},
{
"input": "10\n531086304104565770 690\n831665374188973520 355\n247416398730469667 416\n353278152789250259 494\n679360140254826434 522\n398 867668761409290914\n747 460442302649347038\n428 430798781926999813\n660522762702293186 303\n570848109939810343 23362523897201984\n",
"output": "769690295803806\n2342719363912625\n594750958486742\n715137961111873\n1301456207384766\n2180072264847484\n616388624697949\n1006539210109851\n2179943111228709\n233\n"
},
{
"input": "10\n686 681341882408235553\n961632287055210199 674\n908 102741991216141808\n890 138144729944759257\n419 829534498941150031\n505 624489596307607256\n871 609803422000753610\n983533393633626409 998\n212341556536790997 446\n826239058008214318 864102713930342835\n",
"output": "993209741119930\n1426754135096851\n113151972705234\n155218797690766\n1979795940193742\n1236613061995292\n700118739380921\n985504402438531\n476102144701390\n9639\n"
},
{
"input": "10\n976965156031061996 243\n843847568805985436 621\n573992055384105067 876\n586 308437172337246069\n448791804432291466 662\n34 217523497482835562\n619081882696846159 150\n175189397944495946 977\n654 836460039817622971\n654824899345401176 269167483489754436\n",
"output": "4020432740868597\n1358852767803554\n655242072356317\n526343297503916\n677933239323732\n6397749925965766\n4127212551312329\n179313610997766\n1278990886571336\n415\n"
}
] | 21 | 2 | 256 |
1330_gold_pareidolia
|
1330
|
Pareidolia
| "\nPareidolia is the phenomenon where your eyes tend to see familiar patterns in\nimages where none (...TRUNCATED) |
gold
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1330
|
http://www.usaco.org/index.php?page=open23results
| "\nPareidolia is the phenomenon where your eyes tend to see familiar patterns in\nimages where none (...TRUNCATED) | [{"input":"besssie\n1 1 5 4 6 1 1","output":"1\n4","explanation":"By deleting the 's' at position 4 (...TRUNCATED) | 3 | "s = input()\ncosts = list(map(int, input().split()))\n\ntarget = \"bessie\"\ndp = [(float(\"-inf\")(...TRUNCATED) | "(Analysis by Danny Mittal)\nLet $s$ be the given string, and $c$ be the given list of deletion cost(...TRUNCATED) | [{"input":"besssie\n1 1 5 4 6 1 1\n","output":"1\n4\n"},{"input":"bebesconsiete\n6 5 2 3 6 5 7 9 8 1(...TRUNCATED) | 17 | 2 | 256 |
1326_silver_milk_sum
|
1326
|
Milk Sum
| "\n**Note: The time limit for this problem is 4s, 2x the default.**\nFarmer John's $N$ cows ($1\\le (...TRUNCATED) |
silver
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1326
|
http://www.usaco.org/index.php?page=open23results
| "\n**Note: The time limit for this problem is 4s, 2x the default.**\nFarmer John's $N$ cows ($1\\le (...TRUNCATED) | [{"input":"5\n1 10 4 2 6\n3\n2 1\n2 8\n4 5","output":"55\n81\n98","explanation":"For the first query(...TRUNCATED) | 1 | "\nimport bisect\n\nN = int(input())\narr = list(map(int, input().split()))\nord = sorted(range(N), (...TRUNCATED) | "(Analysis by David Hu)\nNote that it is optimal for Farmer John to milk his cows such that the cow (...TRUNCATED) | [{"input":"5\n1 10 4 2 6\n3\n2 1\n2 8\n4 5\n","output":"55\n81\n98\n"},{"input":"1000\n620 364 41 88(...TRUNCATED) | 11 | 4 | 256 |
1327_silver_field_day
|
1327
|
Field Day
| "\n**Note: The time limit for this problem in Python is 15s. Other languages have the default time (...TRUNCATED) |
silver
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1327
|
http://www.usaco.org/index.php?page=open23results
| "\n**Note: The time limit for this problem in Python is 15s. Other languages have the default time (...TRUNCATED) | [{"input":"5 3\nGHGGH\nGHHHH\nHGHHG","output":"5\n3\n5","explanation":"The first and third teams dif(...TRUNCATED) | 1 | "\nC, N = map(int, input().split())\nto_bin = lambda s: sum(1 << i for i in range(C) if s[i] == \"H\(...TRUNCATED) | "(Analysis by Benjamin Qi)\nA naive solution runs in $O(N^2C)$ time and is too slow to pass any inpu(...TRUNCATED) | [{"input":"5 3\nGHGGH\nGHHHH\nHGHHG\n","output":"5\n3\n5\n"},{"input":"10 100000\nGHGHHHGGHH\nHGHHHH(...TRUNCATED) | 20 | 15 | 256 |
1328_silver_pareidolia
|
1328
|
Pareidolia
| "\n**Note: The time limit for this problem is 4s, 2x the default.**\nPareidolia is the phenomenon wh(...TRUNCATED) |
silver
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1328
|
http://www.usaco.org/index.php?page=open23results
| "\n**Note: The time limit for this problem is 4s, 2x the default.**\nPareidolia is the phenomenon wh(...TRUNCATED) | [{"input":"bessiebessie","output":"14","explanation":"Twelve substrings contain exactly 1 \"bessie\"(...TRUNCATED) | 2 | "\nt = input()\nanswer = 0\nwaiting = [0] * 7\nrem = len(t)\nfor letter in t:\n waiting[0] += 1\n(...TRUNCATED) | "(Analysis by Danny Mittal, Benjamin Qi)\nLet's first consider the easier problem of computing $B(s)(...TRUNCATED) | [{"input":"bessiebessie\n","output":"14\n"},{"input":"abcdefghssijebessie\n","output":"28\n"},{"inpu(...TRUNCATED) | 12 | 4 | 256 |
1325_bronze_rotate_and_shift
|
1325
|
Rotate and Shift
| "\n**Note: The time limit for this problem is 4s, 2x the default.**\nTo celebrate the start of sprin(...TRUNCATED) |
bronze
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1325
|
http://www.usaco.org/index.php?page=open23results
| "\n**Note: The time limit for this problem is 4s, 2x the default.**\nTo celebrate the start of sprin(...TRUNCATED) | [{"input":"5 3 4\n0 2 3","output":"1 2 3 4 0","explanation":"For the example above, here are the cow(...TRUNCATED) | 1 | "\nN, K, T = map(int, input().split())\nA = list(map(int, input().split())) + [N]\n\nans = [-1] * N\(...TRUNCATED) | "(Analysis by Richard Qi)\nFor $N \\le 1000, T \\le 10000$, we can directly simulate the process des(...TRUNCATED) | [{"input":"5 3 4\n0 2 3\n","output":"1 2 3 4 0\n"},{"input":"1000 294 10000\n0 2 6 8 15 17 19 21 24 (...TRUNCATED) | 13 | 4 | 256 |
1302_silver_bakery
|
1302
|
Bakery
| "\nBessie has opened a bakery!\n\nIn her bakery, Bessie has an oven that can produce a cookie in $t_(...TRUNCATED) |
silver
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1302
|
http://www.usaco.org/index.php?page=feb23results
| "\nBessie has opened a bakery!\n\nIn her bakery, Bessie has an oven that can produce a cookie in $t_(...TRUNCATED) | [{"input":"2\n\n3 7 9\n4 3 18\n2 4 19\n1 1 6\n\n5 7 3\n5 9 45\n5 2 31\n6 4 28\n4 1 8\n5 2 22","outpu(...TRUNCATED) | 1 | "\nTC = int(input())\n\nfor tc in range(TC):\n _ = input()\n\n N, X, Y = map(int, input().spli(...TRUNCATED) | "(Analysis by David Hu)\nSuppose Bessie's oven ends up taking $x$ units of time to produce a cookie (...TRUNCATED) | [{"input":"2\n\n3 7 9\n4 3 18\n2 4 19\n1 1 6\n\n5 7 3\n5 9 45\n5 2 31\n6 4 28\n4 1 8\n5 2 22\n","out(...TRUNCATED) | 11 | 2 | 256 |
1303_silver_cow-libi
|
1303
|
Cow-libi
| "\n**Note: The time limit for this problem is 4s, two times the default.**\n\nSomebody has been graz(...TRUNCATED) |
silver
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1303
|
http://www.usaco.org/index.php?page=feb23results
| "\n**Note: The time limit for this problem is 4s, two times the default.**\n\nSomebody has been graz(...TRUNCATED) | [{"input":"2 4\n0 0 100\n50 0 200\n0 50 50\n1000 1000 0\n50 0 200\n10 0 170","output":"2","explanati(...TRUNCATED) | 1 | "\nimport bisect\n\nG, N = map(int, input().split())\n \ndef read():\n x, y, t = map(int, input()(...TRUNCATED) | "(Analysis by Andi Qu)\nFirst, let's think about how to determine if a cow can go from $(x_1, y_1)$ (...TRUNCATED) | [{"input":"2 4\n0 0 100\n50 0 200\n0 50 50\n1000 1000 0\n50 0 200\n10 0 170\n","output":"2\n"},{"inp(...TRUNCATED) | 11 | 4 | 256 |
1299_bronze_hungry_cow
|
1299
|
Hungry Cow
| "\nBessie is a hungry cow. Each day, for dinner, if there is a haybale in the barn,\nshe will eat on(...TRUNCATED) |
bronze
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1299
|
http://www.usaco.org/index.php?page=feb23results
| "\nBessie is a hungry cow. Each day, for dinner, if there is a haybale in the barn,\nshe will eat on(...TRUNCATED) | [{"input":"1 5\n1 2","output":"2","explanation":"","input_explanation":"","output_explanation":""},{(...TRUNCATED) | 3 | "N, T = map(int, input().split())\ndeliveries = [tuple(map(int, input().split())) for _ in range(N)](...TRUNCATED) | "(Analysis by Mihir Singhal)\nWe process the deliveries in time order (as given), keeping track of t(...TRUNCATED) | [{"input":"1 5\n1 2\n","output":"2\n"},{"input":"2 5\n1 2\n5 10\n","output":"3\n"},{"input":"2 5\n1 (...TRUNCATED) | 13 | 2 | 256 |
1300_bronze_stamp_grid
|
1300
|
Stamp Grid
| "\nA stamp painting is a black and white painting on an $N \\times N$ canvas,\nwhere certain cells a(...TRUNCATED) |
bronze
|
http://www.usaco.org/index.php?page=viewproblem2&cpid=1300
|
http://www.usaco.org/index.php?page=feb23results
| "\nA stamp painting is a black and white painting on an $N \\times N$ canvas,\nwhere certain cells a(...TRUNCATED) | [{"input":"4\n\n2\n**\n*.\n1\n*\n\n3\n.**\n.**\n***\n2\n.*\n**\n\n3\n...\n.*.\n...\n3\n.*.\n...\n...(...TRUNCATED) | 1 | "\nT = int(input())\nfor _ in range(T):\n input()\n N = int(input())\n grid = [list(input()(...TRUNCATED) | "\n(Analysis by Claire Zhang) \nConsider stamping wherever possible - that is, for each position and(...TRUNCATED) | [{"input":"4\n\n2\n**\n*.\n1\n*\n\n3\n.**\n.**\n***\n2\n.*\n**\n\n3\n...\n.*.\n...\n3\n.*.\n...\n...(...TRUNCATED) | 14 | 2 | 256 |
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