prob_desc_description stringlengths 63 3.8k | prob_desc_output_spec stringlengths 17 1.47k ⌀ | lang_cluster stringclasses 2
values | src_uid stringlengths 32 32 | code_uid stringlengths 32 32 | lang stringclasses 7
values | prob_desc_output_to stringclasses 3
values | prob_desc_memory_limit stringclasses 19
values | file_name stringclasses 111
values | tags listlengths 0 11 | prob_desc_created_at stringlengths 10 10 | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_notes stringlengths 4 3k ⌀ | exec_outcome stringclasses 1
value | difficulty int64 -1 3.5k ⌀ | prob_desc_input_from stringclasses 3
values | prob_desc_time_limit stringclasses 27
values | prob_desc_input_spec stringlengths 28 2.42k ⌀ | prob_desc_sample_outputs stringlengths 2 796 | source_code stringlengths 42 65.5k | hidden_unit_tests stringclasses 1
value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 178dc786788a3eaa120f056a4e1adbcd | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#include <stdlib.h>
#include <float.h>
#define pi 3.14159265358979323846
int main()
{
int d,h,v,e;
double x,y,z;
scanf("%d %d %d %d",&d,&h,&v,&e);
x=0.25*d*d*pi;
y=v/x;
z=e;
if(y>z){printf("YES\n");
y=y-z;
printf("%.13g\n",h/y);
}
else{pr... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | fad916f9fae0beb7a3ebe2c9f1821eef | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<math.h>
int main(){
double d,h,v,e;
scanf("%lf", &d);
scanf("%lf", &h);
scanf("%lf", &v);
scanf("%lf", &e);
double D = d;
d = d/2;
double vol = M_PI * d*d * h;
double E = e*d*d*M_PI;
double time = vol / (v-E);
//cout<<time<<endl;
if(time <= 10000 && time >= 0){
printf("YES\n... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 0dbf176f8c0f4d2ab7406f083db12924 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#include <math.h>
#define epsilon 0.00000000001
int main() {
int d, h, v, e;
scanf("%d %d %d %d", &d, &h, &v, &e);
double area = d*d*M_PI/4;
double increase_rate = area*e;//*1000;
double volume;
if(increase_rate - v > epsilon) {
printf("NO\n");
}
else {
printf("YES\n");
volume = are... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | c64114587e5560b4aa5af6b834cf3002 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<string.h>
#include<math.h>
const double pi=4*atan(1.0);
int main()
{
int d,h,v,e;
double s,r;
scanf("%d%d%d%d",&d,&h,&v,&e);
r=d/2.0;
if(v<=e*pi*r*r)
printf("NO\n");
else
{
printf("YES\n");
s=(pi*r*r*h/(v-pi*r*r*e));
printf("%.8lf\n",s);
}
return 0;
} | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 3978ed90c460947e05d5738628849e91 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<math.h>
int main(){
double d,h,x,v,t,e;
scanf("%lf%lf%lf%lf",&d,&h,&v,&e);
x=M_PI*d*d*e/4;
if(x<v){
printf("YES\n");
printf("%lf\n",M_PI*d*d*h/4/(v-x));
}
else{
printf("NO\n");
}
return 0;
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 72cff46afcffe4bae65c37697739cced | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
int main()
{
int d,h,v,e,d2,h2;
double pi=acos(-1),t,vol,declevel,uplevel,ans,upvol,downvol;
scanf("%d %d %d %d",&d,&h,&v,&e);
vol=pi*(d/2.0)*(d/2.0)*h;
upvol=pi*(d/2.0)*(d/2.0)*e;
downvol=v;
if(upvol>=downvol)
printf("NO");
else
{
ans=downvol-upvol;
printf("YES\n%.12lf\n",vol/ans);... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 63f9f4bc0c7246a99f170de96f6d932f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
int main ()
{
double d,h,v,e,r,n,x,a,b,c;
double pi =2 * acos(0.0);
scanf("%lf %lf %lf %lf",&d,&h,&v,&e);
r=d/2;
a=pi*r*r*h;
n=pi*r*r*e;
if(v<=n)
printf("NO");
else
{
printf("YES\n");
b=v-n;
c=a/b;
printf("%lf",c);
}}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | dc746580ea1b19119bcb8aeb340f21f7 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#define pi 3.1415926
int main()
{
double d,h,v,e;
while(scanf("%lf%lf%lf%lf",&d,&h,&v,&e)!=EOF)
{
d/=2;
double t=pi*d*d;
t=v/t;
if(t<=e)
printf("NO\n");
else
{
printf("YES\n");
t=h/(t-e);
printf("%lf\n",t);
}
}
return 0;
} | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 65f9ed19252e9b948cc56d989965682a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#define pai 3.14159265359
int main()
{
int i,l,m,n;
double v,e,h,d;
while(scanf("%lf%lf%lf%lf",&d,&h,&v,&e)!=EOF)
{
d=d/2;
v=v/(d*d*pai);
if(e>=v)
printf("NO\n");
else
{
printf("YES\n");
v=v-e;
h=h/v;
printf("%.12lf\n",h);
}
}
} | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 5383ef8f925260e4371aaf21358f4e1d | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define pi 3.14159265
int main() {
double ta;
double icindekisu,hiz,r,R,h,v,e;
scanf("%lf%lf%lf%lf",&R,&h,&v,&e);
r=R/2;
ta=pi*(r*r);
//printf()
icindekisu=ta*h;
hiz=v-(e*ta);
if(hiz<=0){
printf("NO\n");
return 0;
}
if(hiz>0... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | c17cc0f2e9089d7c0d8114d5aedcdbf1 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<math.h>
#include<conio.h>
#define pi 3.14159265358979323846
int main()
{
int a,h,e,v;
double b,g,t;
scanf("%d%d%d%d",&a,&h,&v,&e);
g=pi*a*a*e/4;
if(g<v)
{
b=pi*h*a*a/4;
t=b/(v-g);
printf("\nYES\n%lf",t);
}
else
printf("\nNO");
... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 657aaeb5fc9f27cfd4783e6f83d3c8c8 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<math.h>
int main()
{
int d,h,v,e;
float rate,ans;
scanf("%d%d%d%d",&d,&h,&v,&e);
rate=4*(float)v/((float)(d*d)*M_PI);
//printf("%f",rate);
if(rate<(float)e)
{
printf("NO\n");
}
else
{
printf("YES\n");
ans=(float)h/(rate-(float)e);
printf("%.12f\n",ans);
}
return 0;
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 16773cc8d5cb67672d8ae32669787203 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<malloc.h>
#include<math.h>
#define pi 3.14159265358979323846
int main()
{
int d,v,h,e;
scanf("%d %d %d %d",&d,&h,&v,&e);
double rate=pi*d*d*e/4.0;
rate-=v;
if(rate>=0)
{
printf("NO");
}
else{
double t=(pi*d*d*h)/4.0;
t/=rate;
t*=-1;
printf("YES\n%.12lf",t);
}
retu... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 9dcbf2e1787baf20b1559a33b3f3e3f4 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
int main(void)
{
int d,h,v,e;
double r,A,V,vh,eh,t,pi=3.1415926535;
scanf("%d %d %d %d",&d,&h,&v,&e);
r=d/2.0;
A=pi*r*r;
V=A*h;
vh=v/A;
eh=e*1.0;
if(vh<=eh) {
printf("NO");
}
else {
t=h/(vh-eh);
printf("YES\n%0.6lf",t);
}
return 0;
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | d4a19ac7c361ee7b6f7961e4f17a8279 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#define PI 3.14159265358979323846
int main()
{
double d,h,v,e,area,diff,res;
scanf("%lf %lf %lf %lf",&d,&h,&v,&e);
area=PI*(d/2)*(d/2);
if((v/area)<=e)
{
printf("NO");
}
else
{
printf("YES\n");
diff=(v/area)-e;
res=(1/diff)*h;
pri... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 086360ff81afcd57e35f5452f2f2df67 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#include <math.h>
int main()
{
float r, volagua, volaumenta, res, d, h, e, v;
scanf("%f %f %f %f", &d, &h, &v, &e);
r = d/2;
volagua = r * r * M_PI * h;
volaumenta = r * r * M_PI * e;
if (volaumenta >= v) {
printf("NO");
}
else {
res = volagua * (1/(v - volaumenta));
printf("YE... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 6c1a3d9c4dfb565d0d4202d43321e319 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
int main(void)
{
float pi=3.141592653589;
int d,h,v,e;
scanf("%d%d%d%d",&d,&h,&v,&e);
if(pi*d*d*e/4>=v)
{
printf("NO\n");
}
else
{
printf("YES\n");
printf("%f\n",h*pi*d*d/(v*4-(pi*d*d*e)));
}
return 0;
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 37010ddb9cc46d0ce0621d918159b297 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#define PI 3.1415926535897
double d,h,v,e;
int main()
{
scanf("%lf%lf%lf%lf",&d,&h,&v,&e);
if(PI*(d/2)*(d/2)*e>=v) printf("NO");
else
{
v -= PI*(d/2)*(d/2)*e;
printf("YES\n%.12f",PI*(d/2)*(d/2)*h/v);
}
return 0;
} | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 28a2444de787361d6f24c94cf5f62c16 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
int main()
{
double pi =2 * acos(0.0);
double d,h,v,e;
scanf("%lf %lf %lf %lf",&d,&h,&v,&e);
double a= pi*(d/2)*(d/2);
double g=e*a;
if(g>=v){printf("NO");}
else
{
double ans;
printf("YES\n");
ans=(h*a)/(v-g);
printf("%.12lf",ans);... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 320611a65134338aecc6486dfc487639 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int d,h,v,e;
scanf("%d%d%d%d",&d,&h,&v,&e);
double dij=(double)d/2;
//printf("%lf",dij);
double v2=dij*dij*M_PI*e;
// printf("v2=%lf",v2);
if (v2>=v)
printf("NO\n");
else
{
printf("YES\n");
... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 43f172263241c975920df7bedc2948ad | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
int main()
{
double d,e,h,v;
double p=3.14159265358979,x,t;
scanf("%lf %lf %lf %lf",&d,&h,&v,&e);
x=0.25*p*d*d*e;
t=h/(((4*v)/(p*d*d))-e);
if(x>v)printf("NO");
else printf("YES\n%.12lf",t);
return 0;
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 4dcad9479b02cb89ee91b4fd5e8318e2 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#include <stdlib.h>
#define PI 3.14159265358979323846
int main()
{
double i,j,k,l,m,n,v,h,e,d,val;
double a,s,t;
scanf("%lf %lf %lf %lf",&d,&h,&v,&e);
a=(PI*(d*d))/4;
s=v/a;
if (e>s)
{
printf("NO");
return 0;
}
t=(h)/(s-e);
if (t>10000)
... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 06b9e0ed912fc0b939b76b69d1fe1a10 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#define pi 3.14159265359
int main(){
int d, h, v, e; scanf("%d %d %d %d", &d, &h, &v, &e);
double vol = (pi * d * d * 1.0)/4.0;
double rise = vol * e; //printf("%f ", rise);
double fall = v * 1.0; //printf("%f ", fall);
if(fall > rise){
printf("YES\n");
double ans = (vol * h)/(fall ... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 02ef4b3f298a46778bc0e2b67c9a90db | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#define pi 3.14159265359
int main()
{
long int d,h,v,e;
float x;
scanf("%ld%ld%ld%ld",&d,&h,&v,&e);
x=(float)((v*4)/(pi*d*d))-e;
x<=0?printf("NO"):printf("YES\n%lf",h/x);
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | fc81362bb034e78b0413043cab1d69c0 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <math.h>
#include <stdio.h>
int main() {
int d, h, v, e;
double r, hv;
scanf("%d%d%d%d", &d, &h, &v, &e);
r = (double) d / 2;
hv = (double) v / (r * r * M_PI);
if (e >= hv)
printf("NO\n");
else {
printf("YES\n");
printf("%.12lf\n", h / (hv - e));
}
return 0;
}
| |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | 358d7054ec94a3c48a07ab72dd74321d | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include<stdio.h>
#include<math.h>
int main()
{
long long int d, h, v, e;
double H, t, pi=acos(-1);
scanf("%I64d %I64d %I64d %I64d", &d, &h, &v, &e);
H=(4.0*v)/(pi*d*d);
t=h/(H-e);
if (H<e) printf("NO");
else {
printf("YES\n");
printf("%.12lf", t);
}
... | |
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything ha... | If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is gu... | C | fc37ef81bb36f3ac07ce2c4c3ec10d98 | e6d37a243060f54e2c8074ca7a9c383a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"geometry",
"math"
] | 1461947700 | ["1 2 3 100", "1 1 1 1"] | NoteIn the first example the water fills the cup faster than you can drink from it.In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup wil... | PASSED | 1,100 | standard input | 1 second | The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), where: d — the diameter of your cylindrical cup, h — the initial level of water in the cup, v — the speed of drinking process from the cup in milliliters per second, e — the growth of water because of rain if you do not dri... | ["NO", "YES\n3.659792366325"] | #include <stdio.h>
#define PI 3.14159265359
int main(){
long int d, h, v, e;
scanf("%ld", &d);
scanf("%ld", &h);
scanf("%ld", &v);
scanf("%ld", &e);
if ((v/(PI*((float)d/2)*((float)d/2)))>e){
if(((PI*((float)d/2)*((float)d/2)*h)/(v-e*PI*((float)d/2)*((float)d/2)))<=10000){
p... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | fdc73b37dad79fd3163ad3c788ae6cea | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <stdlib.h>
long long int p[5005]={0};
long long int sum[5005]={0};
long long int f[5005][5005]={0};
long long int max(long long int a,long long int b){
if(a>b)
return a;
else
return b;
}
int main(){
int i,j,n,m,k;
scanf("%d%d%d",&n,&m,&k);
/*int *p = (int *)malloc(n*sizeof(in... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 12443e675c3deb77c3d9bd5e21010515 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
long long int dp[5001][5002];
int main()
{
long long int n,m,k,max=0;
scanf("%lld%lld%lld",&n,&m,&k);
long long int arr[5001],i,a,j;
arr[0]=0;
for(i=1;i<=n;i++)
{
scanf("%lld",&a);
arr[i]=arr[i-1]+a;
}
for(i=0;i<=k;i++)
{
for(j=0;j<=n+1;j++)
... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | b0865ac220555ca71f2cd8fb6058305b | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
#include<stdlib.h>
int a[5100];
long long d[5010][5100];
int main(void)
{
int i,j,p,n,m;
long long sump;
while(scanf("%d%d%d",&n,&m,&p)==3)
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(i=1;i<=n;i++)
{
for(j=1;j<=p;j++)
{
d[i][j]=-1;
}
}
sump=0;
for(j=1;j<=m;j++)
... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | d138a1b2b469e909ebc5ff68827ff8e7 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
#include<stdlib.h>
#define ll long long
ll dp[5005][5005];
ll sum[5005];
int main()
{
int n,m,k,i,j;
ll x,mx;
scanf("%d %d %d",&n,&m,&k);
sum[0]=0;
//memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
scanf("%I64d",&x);
sum[i]=sum[i-1]+x;
}
for(i=0;i<= (... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 1e25b73730cbb42b8c59ba56ea074c71 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
#include<stdlib.h>
#define ll long long
ll dp[5005][2];
ll sum[5005];
ll max(ll a,ll b)
{
return a>b?a:b;
}
int main()
{
int n,m,k,i,j,stat;
ll x,mx;
scanf("%d %d %d",&n,&m,&k);
sum[0]=0;
//memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
scanf("%I64d",&x);
... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 47ee725889aa42bbb42aee3a8bd244a8 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <string.h>
#define ll long long
#define MAXN 5050
ll dp[MAXN][MAXN], p[MAXN], c[MAXN];
ll imax(ll a, ll b){
return a > b ? a : b;
}
int main(){
int i, j, n, m, k;
while(scanf("%d %d %d", &n, &m, &k) != EOF){
c[0] = 0;
for(i = 1; i <= n; i++){
scan... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | af2ea1880623c4fe574349431e092f9f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
long long dp[5005][5005];
long long A[5005];
long long pre[5005];
long long max(long long a,long long b)
{
if(a>b)
return a;
else
return b;
}
long long rec(long long start,long long int m,long long int n,long long int k)
{
//when you are at the start o... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 38eda133bf8a44f875380bd05b09aff9 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
#include<stdbool.h>
long long int max[5001][5001],sum2[5001];
bool yes[5001][5001];
int n,m,k;
long long int dp(int start,int k2)
{
if(k2==0||n-start<m*k2)
return 0;
if(yes[start][k2]==1)
return max[start][k2];
int i;
long long int max2,a=dp(start+1,k2... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | a8598c77b9b3aaa554f1c26f2b79ac42 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
#define INF -6e12
typedef long long int LL;
int n,m;
LL sum[5005];
LL arr[5005];
LL memo[5005][5005];
char calculated[5005][5005];
LL MAX(LL a,LL b)
{
return a > b ? a : b;
}
LL dp(int index,int k)
{
// Meaning that we have crossed the array boundary but we still have some
// m sized ranges ... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | f2bfb12aa5f93768a58405b21e0baa46 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
#define LL long long
LL sum[5010];
LL dp[5010][5010];
LL max(LL a,LL b)
{
return a>b?a:b;
}
int main()
{
int n,k,m,i,num,j;
scanf("%d %d %d",&n,&m,&k);
sum[0]=0;
for(i=1;i<=n;++i)
{
scanf("%d",&num);
sum[i]=sum[i-1]+num;
}
for(i=m;i<=n;++i)
{
... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 3b5d73412b0af8def403d0968b063a1d | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
int main()
{
long long int n,i,j,k,h,t,m;
long long int prev[5001],cur[5001];
long long int b[5001],sum[5001],count;
scanf("%lld %lld %lld",&n,&m,&k);
for(i=0;i<n;i++)
scanf("%lld",&b[i]);
for(i=0;i<n;i++)
prev[i]=0;
count=0;
if(n==1)
printf("%lld\n",b[0]);
else
{
for(i=0;i<m-1;i++)
{
sum[i]=0;
count+... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | f533afa7b363ef5fd9ed21cbe334ca60 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <stdlib.h>
int main()
{
int no,m,k,i,j;
scanf("%d %d %d",&no,&m,&k);
int a[5002]={0};
long long dp[5001][5001]={0};
long long sum[5002]={0};
for(i=1;i<=no;i++)
scanf("%d",&a[i]);
long long sum1=0;
j=1;
for(i=no;j<=m;i--)
{sum1=sum1+a[i];j++;}/... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 895e252cd5b6922508d23af8d435ba1a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include<stdio.h>
long long int array[5001][5001];
long long int max(long long int a,long long int b)
{ if(a>b)
return a;
else
return b;
}
long long int railway(long long int b[],long long int index,long long int k,long long int m)
{ if(k==0|| index<0)
return 0 ;
if(array[index][... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 1fbcd971de226279345a7fdab17f02be | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <stdlib.h>
#define max(a,b) (a>b?a:b)
int main(){
long temp[5010];
long long a[5010], arr[5010][5010];
int n,m,k,i,j;
while(scanf("%d%d%d" , &n,&m,&k) != EOF){
int ans = 0;
a[0] = 0;
for(i = 1; i <= n; i++) {
scanf("%ld", &temp[i]);
a[i] = a[i - 1] +... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | b10bdb287cee3d59c5345c54c437b69a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <stdlib.h>
#define max(a,b) (a>b?a:b)
int main(){
long a[5010];
long long s[5010], dp[5010][5010];
int n,m,k,i,j;
while(scanf("%d%d%d" , &n,&m,&k) != EOF){
int ans = 0;
s[0] = 0;
for(i = 1; i <= n; i++) {
scanf("%ld", &a[i]);
s[i] = s[i - 1] + a[i];
... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | ba4de93e99194d06e2008ad2d04d13f0 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
long long n,m,k;
long long sum[5010];
long long save[5010][5010];
long long ans = 0;
long long max(long long a, long long b)
{
return (a>b)?a:b;
}
long long bf(long long l, long long ind)
{
long long r = m - 1 + l;
if (ind > k)
return 0;... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | eb01596e7c9017455a0781fe1cdf6f11 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | #include <stdio.h>
#define max(x, y) ((x)>(y)?(x):(y))
long long P[5001];
long long Dp[5001][5001];
long long Ans;
int main()
{
int N, M, K, i, j;
scanf("%d %d %d", &N, &M, &K);
for(i=1; i<=N; ++i)
{
scanf("%d", &P[i]);
P[i] += P[i-1];
}
for(i=1; i<=N; ++i)
{
for(j=... | |
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers: [l1, r1], [l2, r... | Print an integer in a single line — the maximum possible value of sum. | C | ee3c228cc817536bf6c10ea4508d786f | 45bf1f118b582686b9e49ae3dffeab33 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"dp",
"implementation"
] | 1411054200 | ["5 2 1\n1 2 3 4 5", "7 1 3\n2 10 7 18 5 33 0"] | null | PASSED | 1,700 | standard input | 1 second | The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109). | ["9", "61"] | /*
* 1.这个程序做到的事情
* 2.这个程序遇到的困难
* 我改进的一个方向就是！
* 3.接下来要做的事情
*/
#include<stdio.h>
#define N 5002//我没有注意，把这一行注视了，用了下一行！
//#define N 502
long long max2(long long a,long long b){
if(a>b)return a;
else return b;
}
int main() {
int n,m,k;
int p[N];
scanf("%d%d%d",&n,&m,&k);
/*
printf("n=%d\n",n... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | f508c4511322afd9151122cdfe559da5 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<string.h>
int main()
{
int n;
scanf("%d", &n);
int i, j;
char s[200005];
scanf("%s", s);
int ans[50004];
int m;
int l;
char t[200005];
int count[30][200005];
for (i = 0; i < 30; i++)
count[i][0] = 0;
int min, max, mid;
int x;
for (i = 0; i < n; i++)
{
for (j = 0; j < 30; j+... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 5b9e1197859d9f82fb60f07a21b229a6 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<string.h>
int test[126];
void clear()
{
int i;
for(i=97;i<=125;i++)
test[i]=0;
}
int main()
{
int n,i,m,j,temp,max=0,len,k,cnt[126]={};
scanf("%d",&n);
char str[n+2],name[n+2];
int value[n+2][125];
scanf("%s",str);
for(i=0;str[i]!='\0';i++)
{
... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | efc163911bda48963cbc34688f2ed1cb | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | //practice with dukkha
#include<stdio.h>
#include<string.h>
#define mod 1000000007
int com[26][2000000];
int comp[26];
int main()
{
int a;
scanf("%d",&a);
char array[a];
scanf("%s",array);
for(int i=0;i<a;i++)
{
com[array[i]-'a'][comp[array[i]-'a']]=i;
comp[array[i]-'a']++;
... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | b610c91ecfb339334f5b65604dce7437 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX_LEN 200001
int main() {
int shoplen;
scanf("%d", &shoplen);
char shop[shoplen + 1];
scanf("%s", shop);
int *count = calloc(26, sizeof(int));
int **posmat = calloc(26, sizeof(int*)) ;
int sum = 0 ;
posmat[0] = calloc(shoplen, size... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 825380b46bd5f0570ce280621f2da03f | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
#include <string.h>
#define MAX_LEN 200001
int main() {
int shoplen;
scanf("%d", &shoplen);
char shop[shoplen + 1];
scanf("%s", shop);
// int posmat[26][shoplen];
int shopcount[26];
int *count = calloc(26, sizeof(int));
int **posmat = calloc(26, sizeof(int*)) ;
int sum = 0 ;
p... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 45edc830e4ee6369259bfad0da40b21a | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
// #include <stdlib.h>
#include <string.h>
#define MAX_LEN 200001
int main() {
int shoplen;
scanf("%d", &shoplen);
char shop[shoplen + 1];
scanf("%s", shop);
int posmat[26][shoplen];
int shopcount[26];
for (int x = 0; x < 26; x++) {
for (int y = 0; y < shoplen; y++) {
posm... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | f1ad4bb928236d44b507fccee0ffe929 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
#include <string.h>
int chars_s[26][200002];
char s[200001];
char t[200001];
int main()
{
int n;
scanf("%d", &n);
scanf("%s", s);
int chars_s_index[26];
for(int i = 0; i < 26; i++)
{
chars_s_index[i] = 0;
chars_s[i][0] = 0;
}
int len_s = n;
for(int i = 0; i < len_s; i++)
chars_s[s[i]... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | c3ec56d4d31fa61659c3ef0f328deea5 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
int main(){
int n,n1,i,count=0,j,k,k1,a[26]={0},max=0,l,c[26]={0};
scanf("%d",&n);
getchar();
char s[n];
gets(s);
int b[26][200001]={0};
for(k=0;s[k]!=NULL;k++){
a[s[k]-97]++;
b[s[k]-97][a[s[k]-97]-1]=(k+1);
}
scanf("%d",&n1);
getchar();
char s1[300000];
for(j=0;j<n... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 05c9bb27b7a291fcfae5fb5a375d3d79 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
int main(){
int n,n1,i,count=0,j,k,k1,a[26]={0},max=0,l,c[26]={0};
scanf("%d",&n);
getchar();
char s[n];
gets(s);
int b[26][200001]={0};
for(k=0;s[k]!=NULL;k++){
a[s[k]-97]++;
b[s[k]-97][a[s[k]-97]-1]=(k+1);
}
scanf("%d",&n1);
getchar();
char s1[300000];
for(j=0;j<n1... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 13410dbca9ddc1ed94bcfebe0d11b203 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
int n,q;
char s[200010],t[200010];
int cnts[200010][30],cntt[30];
short check(int x)
{
int i;
for(i=0;i<26;i++) if(cnts[x][i] < cntt[i]) return 0;
return 1;
}
int work()
{
int i,head = 1,tail = n,mid;
for(i=0;i<26;i++) cntt[i] = 0;
for(i=1;t[i];i++) cntt[t[i]-'a']++;
... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 463ecef515d31342d4c667564d999476 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | /* practice with Dukkha */
#include <stdio.h>
#include <string.h>
#define N 200000
#define A 26
int main() {
static char aa[N + 1], bb[N + 1];
static int kk[N][A], ll[A];
int n, q, i, a;
scanf("%d%s", &n, aa);
for (i = 0; i < n; i++)
for (a = 0; a < A; a++)
kk[i][a] = (i > 0 ? kk[i - 1][a] : 0) + (aa[i] ==... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | ebe7c1d39441b9022dbcb6f690719aba | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | /* Coached by rainboy */
#include <stdio.h>
#include <string.h>
#define N 200000
int main() {
static char s[N + 1];
static int kk[N][26], ll[26];
int n, q, i;
scanf("%d%s%d", &n, s, &q);
for (i = 0; i < n; i++) {
ll[s[i] - 'a']++;
memcpy(kk[i], ll, 26 * sizeof *ll);
}
while (q-- > 0) {
static char t[N +... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 23ef3dc69aa9673b878598760aeb458d | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
char s[200005];
int cnt[26][200005] = { 0 };
char fs[200005];
int fcnt[26] = { 0 };
#define MAX(a,b) ((a)>(b)?(a):(b))
int main()
{
int len;
scanf("%d", &len);
scanf("%s", s);
for (int i = 0; i < len; i++)
{
cnt[s[i] - 'a'][i]++;
//转移
for (int c = 0; c ... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 5774bc5885ca96fe600d295f174c8435 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
#include <string.h>
#define N 200000
int aa[28],bb[28];
int ss[28][N];
char s2[N];
int main(){
int n,a;
char x;
scanf("%d",&n);
char c=getchar();
for(int i=0;i<n;i++){
scanf("%c",&x);
a=(int) x-97;
aa[a]++;
ss[a][aa[a]]=i;
}
int m;
scanf("%d",&m);
c=getchar();
while(m--){
for(in... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 7ba5d28a40a6f941df45497bc513dabc | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
long long m, k, l, t, i, j, k, a[30], c[30][300000], d, x, y, b, g, w, e;
char s[1000000];
int war(const void* aa, const void* bb)
{
return(*(long long*)aa-*(long long*)bb);
}
/*
long long bins(long long p, long long k, long long s)
{
if(a[(p+k)/2]>=... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 3da4467bee358c5f5555bdd619457bb9 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include <stdio.h>
#include <string.h>
int chars_s[26][200002];
char s[200001];
char t[200001];
int main()
{
int n;
scanf("%d", &n);
scanf("%s", s);
int chars_s_index[26];
for(int i = 0; i < 26; i++)
{
chars_s_index[i] = 0;
chars_s[i][0] = 0;
}
int len_s = n;
for(int i = 0; i < len_s; i++)
chars_s[s[i]... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 299666a28105941ee13bda687ff7f8d5 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>
typedef int64_t i64;
typedef int32_t i32;
static void print_int(i64 n){if(n<0){putchar('-');n=-n;}if(n==0){putchar('0');return;}int s[20],len=0;while(n>0){s[len++]=n%10+'0';n/=10;}while(len>0){putchar(s[--len]);}}
static i64 read_int(void){i... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 58641df38b0cae477721aa4c695567d5 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 200002
int main(){
int n;
scanf("%d",&n);
char s[N];
scanf("%s",s);
int ns=strlen(s);
//printf("s=%s\n",s);
//printf("ns=%d\n",ns);
int scount[26];
int i,j;
for(i=0;i<26;i++){
scount[i]=0;
}
for(i=0... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | c39a3e0a6066303c930bfff35df6dff9 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<string.h>
#include <stdlib.h>
#include <conio.h>
#define debug printf("SUSU\n")
int latter[26][300000];
int main()
{
int n,m,i,j;
char input[300000];
int cnt[300000]={};
char a;
scanf("%d ",&n);
for(i=0;i<n;i++)
{
scanf("%c",&a);
latter[a-'a'][cnt... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | eec2cdabcd5d5d0ccc120da9d2b9002a | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<string.h>
int e[27][200009];
int main()
{
int n,m,i,j,k,f[30]={},count,g,h[30]={},l,z;
char c[200009],d[20009];
scanf("%d",&n);
scanf("%s",&c);
getchar();
for(i=0;i<n;i++){
g=c[i]-96;
h[g]++;
e[g][h[g]]=i;
}
s... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 1b64535ababc7a4820c3f702cd033805 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<string.h>
typedef long long int ll;
int max(ll x,ll y)
{
return x>y?x:y;
}
int main()
{
ll n,i;
char s[1000000];
scanf("%lld",&n);
scanf("%s",s);
ll arr[26][1000000];
ll ind[26];
for (i=0;i<26;i++) ind[i]=0;
for (i=0;i<n;i++)
{
arr[s[i]-'a'][ind[s[i]-'a']]=i+1;
ind[s[i]-'a']++;
... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 895261178cd57a770ca6a5fee4653645 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
int main(){
int n,n1,i,count=0,j,k,k1,a[26]={0},max=0,l,c[26]={0};
scanf("%d",&n);
getchar();
char s[n];
gets(s);
int b[26][200001]={0};
for(k=0;s[k]!=NULL;k++){
a[s[k]-97]++;
b[s[k]-97][a[s[k]-97]-1]=(k+1);
}
scanf("%d",&n1);
getchar();
char s1[300000];
for(j=0;j<n... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | 07090e86843cf04160332e167e97122a | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
int aaa[26][300000];
int aa[26];
int main()
{
int a,t,i,c,j;
scanf("%d ",&a);
char b[a];
for(i=0;i<a;i++)
{
scanf("%c",&b[i]);
//b[i]=b[i]-';
aaa[b[i]-'a'][aa[b[i]-'a']]=i+1;
// printf("%d=%d\n",aa[b[i]-'a'],aaa[b[i]-'a'][aa[b[i]-'a']]);
aa[b[... | |
The letters shop showcase is a string $$$s$$$, consisting of $$$n$$$ lowercase Latin letters. As the name tells, letters are sold in the shop.Letters are sold one by one from the leftmost to the rightmost. Any customer can only buy some prefix of letters from the string $$$s$$$.There are $$$m$$$ friends, the $$$i$$$-th... | For each friend print the length of the shortest prefix of letters from $$$s$$$ s/he would need to buy to be able to construct her/his name of them. The name can be constructed if each letter is presented in the equal or greater amount. It is guaranteed that every friend can construct her/his name using the letters fro... | C | 8736df815ea0fdf390cc8d500758bf84 | f0fa201cef501c1fa0769ba1a1430400 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"binary search",
"implementation",
"strings"
] | 1561905900 | ["9\narrayhead\n5\narya\nharry\nray\nr\nareahydra"] | null | PASSED | 1,300 | standard input | 2 seconds | The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of showcase string $$$s$$$. The second line contains string $$$s$$$, consisting of exactly $$$n$$$ lowercase Latin letters. The third line contains one integer $$$m$$$ ($$$1 \le m \le 5 \cdot 10^4$$$) — the number of friends. The ... | ["5\n6\n5\n2\n9"] | #include<stdio.h>
#include<string.h>
int test[126];
void clear()
{
int i;
for(i=97;i<=125;i++)
test[i]=0;
}
int main()
{
int n,i,m,j,temp,max=0,len,k,cnt[126]={};
scanf("%d",&n);
char str[n+2],name[n+2];
int value[n+2][125];
scanf("%s",str);
for(i=0;str[i]!='\0';i++)
{
... | |
You are given n points on Cartesian plane. Every point is a lattice point (i. e. both of its coordinates are integers), and all points are distinct.You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines? | If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO. | C | a9fd2e4bc5528a34f1f1b869cd391d71 | e9d564f5041b4069c1de67662696162a | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"geometry"
] | 1522850700 | ["5\n0 0\n0 1\n1 1\n1 -1\n2 2", "5\n0 0\n1 0\n2 1\n1 1\n2 3"] | NoteIn the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points. | PASSED | 2,000 | standard input | 2 seconds | The first line contains one integer n (1 ≤ n ≤ 105) — the number of points you are given. Then n lines follow, each line containing two integers xi and yi (|xi|, |yi| ≤ 109)— coordinates of i-th point. All n points are distinct. | ["YES", "NO"] | #include<stdio.h>
int gcd(int a,int b)
{
int c;
while(a%b!=0)
{
c=b;
b=a%b;
a=c;
}
return b;
}
int main()
{
int n,i,j,k,con,refx,refy,delx1,dely1,dely2,delx2,lastx,sign;
sign=con=1;
scanf("%d",&n);
int p[n][3];
for(i=0;i<n;i++)
{
scanf("%d %d",... | |
You are given n points on Cartesian plane. Every point is a lattice point (i. e. both of its coordinates are integers), and all points are distinct.You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines? | If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO. | C | a9fd2e4bc5528a34f1f1b869cd391d71 | 16e347efa6dc3797e603f3d189fa98cf | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"geometry"
] | 1522850700 | ["5\n0 0\n0 1\n1 1\n1 -1\n2 2", "5\n0 0\n1 0\n2 1\n1 1\n2 3"] | NoteIn the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points. | PASSED | 2,000 | standard input | 2 seconds | The first line contains one integer n (1 ≤ n ≤ 105) — the number of points you are given. Then n lines follow, each line containing two integers xi and yi (|xi|, |yi| ≤ 109)— coordinates of i-th point. All n points are distinct. | ["YES", "NO"] | //set many funcs template
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384
int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a... | |
Furlo and Rublo play a game. The table has n piles of coins lying on it, the i-th pile has ai coins. Furlo and Rublo move in turns, Furlo moves first. In one move you are allowed to: choose some pile, let's denote the current number of coins in it as x; choose some integer y (0 ≤ y < x; x1 / 4 ≤ y ≤ x1 / 2) and de... | If both players play optimally well and Furlo wins, print "Furlo", otherwise print "Rublo". Print the answers without the quotes. | C | cc23f07b6539abbded7e120793bef6a7 | e1cce01a825a4a58eec4f89ff41052c0 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"games"
] | 1355671800 | ["1\n1", "2\n1 2", "10\n1 2 3 4 5 6 7 8 9 10"] | null | PASSED | 2,200 | standard input | 2 seconds | The first line contains integer n (1 ≤ n ≤ 77777) — the number of piles. The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 777777777777) — the sizes of piles. The numbers are separated by single spaces. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the c... | ["Rublo", "Rublo", "Furlo"] | /* practice with Dukkha */
#include <math.h>
#include <stdio.h>
#define A 888888
int tr[A + A];
void update(int i, int b) {
tr[i += A] = b;
while (i > 1) {
i >>= 1;
tr[i] = tr[i << 1] | tr[i << 1 | 1];
}
}
int query(int l, int r) {
int b = 0;
for (l += A, r += A; l <= r; l >>= 1, r >>= 1) {
if ((l & 1) ... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 1ef61f48a52b76f86ffd9a10333f7928 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
#include<string.h>
int GY(int a, int b)
{
if(b == 0)
return a;
if(a % b == 0)
return b;
else
return GY(b, a % b);
}
int main()
{
int a, b, c[200], d[200], m, n, x, y, i, j, k, t;
while(scanf("%d%d", &m, &n) != EOF)
{
memset(c, 0, sizeof(c));
memset(d, 0, sizeof(d));
scanf("%d", &a);... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 42131ed3f9e0ee50f05d4d7a146f9869 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
int main()
{
int g,i,j,k,l,m,n,a,b,s;
scanf("%d",&n);
scanf("%d",&m);
short int B[100],G[100];
for(i=0;i<100;i++)
{
B[i]=0;
G[i]=0;
}
scanf("%d",&b);
for(i=0;i<b;i++)
{
scanf("%d",&j);
B[j]=1;
}
scanf("%d",&g);
for(... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | c82ccca2fe46ffc1711f5012d990f0bb | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#include <stdlib.h>
int boys[100]={0};
int girls[100]={0};
int main()
{
int b,g,hb,hg,i,j,x,y;
scanf("%d %d",&b,&g);
scanf("%d",&hb);
for(i=0;i<hb;i++){
scanf(" %d",&x);
boys[x]=1;
}
scanf("%d",&hg);
for(i=0;i<hg;i++){
scanf(" %d",&x);
girls... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | ac5e5ee0b81fab3757e480a4f56b4d53 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
#include<stdlib.h>
int
main(void){
int m,n,i,j,input,b,g;
scanf("%d %d",&n,&m);
scanf("%d",&b);
int *b_list = (int *)calloc(n,sizeof(int));
for(i=0;i<b;i++){
scanf("%d",&input);
b_list[input]=1;
}
scanf("%d",&g);
int *g_list = (int *)calloc(m,sizeof(i... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 36cf5dc982513e93301af37ebf625be9 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
int x[100], y[100];
int main(void)
{
int i, j, n, m, b, g, last = -1;
scanf("%d %d %d", &n, &m, &b);
for (i = 0; i < b; i++) {
scanf("%d", &j);
x[j] = 1;
}
scanf("%d", &g);
for (i = 0; i < g; i++) {
scanf("%d", &j);
y[j] = 1;
}
for (i ... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 4e752a2670fc98cd5eb6c55297b46226 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
#include<string.h>
int b[10010], g[10010];
int vis[110];
int que[10010];
int main()
{
int m, n, h, sub;
int i, j, cont = 0;
memset(vis, 0, sizeof(vis));
scanf("%d %d", &n, &m);
for (i = 0; i < n*m; i++)
b[i] = i%n;
for (i = 0; i < n*m; i++)
g[i] = i%m;
scanf("%d", &h);
for (i = 0; i ... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 3f27f88c63509a6a5bd25ee29c4b3b02 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
int main()
{
int n,m,b,g,x,i,j;
int B[100]={0},G[100]={0};
scanf("%d%d",&n,&m);
scanf("%d",&b);
for(i=0;i<b;i++)
{
scanf("%d",&x);
B[x]=1;
}
scanf("%d",&g);
for(i=0;i<g;i++)
{
scanf("%d",&x);
G[x]=1;
}
for(i=0;i<m*n+m+n;i... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 940c653269762d68894c51f452e663b2 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
int max(int a, int b)
{
if (a>b) return a;
else return b;
}
int min(int a, int b)
{
if (a<b) return a;
else return b;
}
int gcd(int a, int b){
int c, d;
d = max(a, b); c = min(a, b);
if(d%c == 0) return c;
else return gcd(d%c, c);
}
int p... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 45dfa2de3a88013dda278e920089e7aa | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
#include<string.h>
int gcd(int a, int b) {
if(a > b) return gcd(b, a);
if(b % a == 0) return a;
return gcd(b % a, a);
}
int main()
{
int n,m,i,j,x,flag,sign=0;
int a[105],b[105],tmp[105];
memset(a,-1,sizeof(a));
memset(b,-1,sizeof(b));
memset(tmp,0,sizeof(tmp));
scanf("%d %d",&n,&m);... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 051b908ba4aa5ad1f3878033023654e7 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
int max(int a, int b)
{
if (a>b) return a;
else return b;
}
int min(int a, int b)
{
if (a<b) return a;
else return b;
}
int gcd(int a, int b){
int c, d;
d = max(a, b); c = min(a, b);
if(d%c == 0) return c;
else return gcd(d%c, c);
}
int p... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | afd7f05b57d8d1ced55d85799078b404 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
int n;
int left,right,middle;
int a[10000],b[10000];
void msort(int a[],int left,int middle,int right)
{
int temp,temp1,temp2,l1,l2,d,d1,t1,flag,i;
temp1=middle+1;
temp=left;
temp2=right;
l1=middle-left+1;
l2=right-middle; d=l1; d1=l2; t1=left;
flag=0;
while(d!=0&&d1!=0)
{
if(a[temp]<... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 9b094298ab5117b26645286ece67b77d | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#define maxn 120
int main()
{
int n,i,j,m,b,g;
int x[maxn]={0},y[maxn]={0};
scanf("%d%d",&n,&m);
scanf("%d",&b);
for (i=0;i<b;i++){
scanf("%d",&j);
x[j]=1;
}
scanf("%d",&g);
for (i=0;i<g;i++){
scanf("%d",&j);
y[j]=1;
}
for (i=0;i<=4*n*m;i++)
if (x[i%n]||y[i%m]) {x[i%n]=1;y[i%m]=1;... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 7f04722ac82c72158d9a197afbd88d93 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#include <stdlib.h>
int main(void)
{
int n, m;
scanf("%d %d", &n, &m);
int i, j;
int boy[n];
int girl[m];
for (i = 0; i < n; ++i) {
boy[i] = 0;
}
for (i = 0; i < m; ++i) {
girl[i] = 0;
}
int b;
scanf("%d", &b);
int temp;
for (i = 0... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | c769848b0fcc2356c6915040670f8916 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
int main(void) {
int n,m,b,g,i,index,ans1,ans2;
int flag1=0;
int flag2=0;
scanf("%d%d",&n,&m);
int b1[n];
int g1[m];
for(i=0;i<n;i++)b1[i]=0;
for(i=0;i<m;i++)g1[i]=0;
scanf("%d",&b);
for(i=0;i<b;i++)
{
scanf("%d",&index);
b1[index]=1;
}
... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 356ce739bc4f7448b1402cab978cbe82 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
int arr[2500];
int arr1[2500];
int max(int a,int b){
return a>b?a:b;
}
int main(){
int a,b;
scanf("%d%d",&a,&b);
int i; int left,left1;
int x,g;
scanf("%d",&x);
for(i=0;i<x;i++){
int in;
scanf("%d",&in);
arr[in]=1;
}
scanf("%d",&g);
for(i=0;i<g;i++){
int in;
scanf("%d",&in);
arr1[in]=1... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 8fccbc424c6f27b52b060effdc47036d | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#include <string.h>
int gcd(int x,int y)
{
if (x%y==0)
return y;
else
if (y%x==0)
return x;
else
if (x>y)
return gcd(y,x%y);
else
return gcd(x,y%x);
}
int main()
{
int n,m,b,g,x[100],d,i,j;
for (i=0;i<100;i++) x[i]=0;
scanf(" %d %d",&n,&m);
d=gcd(n,m);
scanf(" %d",&b);... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 0fee90a71db45f7c3cd908aadcbe75f5 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
#include <string.h>
void swap(int *x,int *y)
{
int k=*x;*x=*y;*y=k;
}
int gcd(int x,int y)
{
if (y>x) swap(&x,&y);
if (x%y==0)
return y;
else
return gcd(y,x%y);
}
int main()
{
int n,m,b,g,x[100],d,i,j;
for (i=0;i<100;i++) x[i]=0;
scanf(" %d %d",&n,&m);
d=gcd(n,m);
scanf(" %d",&b);
fo... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 780311e45066a69fa07d11e8be039db7 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
typedef unsigned u;
u A[100],B[100];
int main()
{
u n,m,i,j,N,x,y;
scanf("%u%u",&n,&m);
for(scanf("%u",&x);x--;)
{scanf("%u",&y);A[y]=1;}
for(scanf("%u",&x);x--;)
{scanf("%u",&y);B[y]=1;}
while(getchar()==' ');
for(N=n*m,i=1,x=y=0;i;)for(i=j=0;j<N;++j)
{
if(A[x]^B[y])A[x]=B[y]=i=1;
if(++x... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 960d22144342f7170489265b31578dd9 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | /* .................... compiled by alankar....................
*/
//......................SHORTCUTS..............................
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
//...........................................................
#define pi 3.14159265358979323846 //float type
#de... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | be4d94179bfb20501335099174eea801 | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
int main()
{
int arr1[1000]={0};
int arr2[1000]={0};
int n,m;
scanf("%d %d",&n,&m);
int b;
scanf("%d",&b);
int i;
int x;
for(i=0;i<b;i++)
{
scanf("%d",&x);
arr1[x]=1;
}
int g... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 933f2f6ff94d9548afa21759767aaf1c | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include<stdio.h>
int main(){
int flag,head1,head2,sum,i,j,k,l,n,m,bn,gn,b[2][1000],g[2][1000];
scanf("%d%d",&n,&m);
scanf("%d",&bn);
for(i=0;i<bn;i++){
scanf("%d",&k);
b[1][k]=1;
}
scanf("%d",&gn);
for(i=0;i<gn;i++){
scanf("%d",&k);
g[1][k]=1;
}
flag=1,head1=0,head2=0,sum=0,head1=0,head2=0;
while(sum... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 946a60e18a9d13eb67bd95f9f8b9786b | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdlib.h>
int getint(){
char c,sign=1;
int ans=0;
while((c=getchar())<'-');
if(c=='-')sign=-1,c=getchar();
do{
ans=ans*10+c-48;
}while((c=getchar())>='0');
return ans*sign;
}
int main(){
int n, m, a, b, x;
int i;
char boy[100]={0}, girl[100]={0};
n=getint(),m=getint();
a=getint();
for(i=0;i<a... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | e40c8bc74c10139a3e30ca801183b2aa | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
int main()
{
int boy[100], girl[100], happy_boys, happy_girls;
int n, m, i, flag, temp, count=10001;
int counter_boy=0, counter_girl=0;
scanf("%d %d", &n, &m);
for(i=0; i<100; i++) {
boy[i] = 0;
girl[i] = 0;
}
scanf("%d", &happy_boys);
for (i=0; i <... | |
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites -th boy and -th girl... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | C | 65efbc0a1ad82436100eea7a2378d4c2 | 5c04fa5b5b8c69333c74b2a6dee3a58a | GNU C | standard output | 256 megabytes | train_001.jsonl | [
"dsu",
"meet-in-the-middle",
"number theory",
"brute force"
] | 1424190900 | ["2 3\n0\n1 0", "2 4\n1 0\n1 2", "2 3\n1 0\n1 1"] | NoteBy we define the remainder of integer division of i by k.In first sample case: On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes... | PASSED | 1,300 | standard input | 2 seconds | The first line contains two integer n and m (1 ≤ n, m ≤ 100). The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys. The third line conatins integer g (0 ≤... | ["Yes", "No", "Yes"] | #include <stdio.h>
int main()
{
int boy[100], girl[100], happy_boys, happy_girls;
int n, m, i, flag, temp, count=10001;
int counter_boy=0, counter_girl=0;
scanf("%d %d", &n, &m);
for(i=0; i<100; i++) {
boy[i] = 0;
girl[i] = 0;
}
scanf("%d", &happy_boys);
for (i=0; i <... | |
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form .Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can't ... | If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1. If there are multiple answers, print any of them. | C | f60ea0f2caaec16894e84ba87f90c061 | 85e9199263adc09a3112f48e8ae918bd | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"constructive algorithms",
"number theory",
"brute force",
"math"
] | 1481726100 | ["3", "7"] | null | PASSED | 1,500 | standard input | 1 second | The single line contains single integer n (1 ≤ n ≤ 104). | ["2 7 42", "7 8 56"] | #include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
if(n-1) printf("%d %d %d\n",n,n+1,(n*(n+1)));
else printf("-1\n");
return 0;
}
| |
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form .Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can't ... | If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1. If there are multiple answers, print any of them. | C | f60ea0f2caaec16894e84ba87f90c061 | c7e4db1cf6e6b8383756f754a931b683 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"constructive algorithms",
"number theory",
"brute force",
"math"
] | 1481726100 | ["3", "7"] | null | PASSED | 1,500 | standard input | 1 second | The single line contains single integer n (1 ≤ n ≤ 104). | ["2 7 42", "7 8 56"] | #include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int n;
scanf("%d",&n);
if(n==1)
printf("-1\n");
else
{
printf("%d ",n);
printf("%d ",n+1);
printf("%d\n",n*(n+1));
}
return 0;
}
| |
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form .Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can't ... | If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1. If there are multiple answers, print any of them. | C | f60ea0f2caaec16894e84ba87f90c061 | 3464b54caff1f7b44e7262e4e002e853 | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"constructive algorithms",
"number theory",
"brute force",
"math"
] | 1481726100 | ["3", "7"] | null | PASSED | 1,500 | standard input | 1 second | The single line contains single integer n (1 ≤ n ≤ 104). | ["2 7 42", "7 8 56"] | #include <stdio.h>
int main()
{
long int n;
scanf("%d",&n);
if(n!=1)printf("%ld %ld %ld",n,n+1,n*(n+1));
else printf("%ld",-1);
return 0;
}
| |
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form .Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can't ... | If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1. If there are multiple answers, print any of them. | C | f60ea0f2caaec16894e84ba87f90c061 | 0076080de17603701b6f51c13c616c0a | GNU C11 | standard output | 256 megabytes | train_001.jsonl | [
"constructive algorithms",
"number theory",
"brute force",
"math"
] | 1481726100 | ["3", "7"] | null | PASSED | 1,500 | standard input | 1 second | The single line contains single integer n (1 ≤ n ≤ 104). | ["2 7 42", "7 8 56"] | #include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<limits.h>
#define MOD 1000000007
#define PI 3.14159265
#define seive_len 1000001
int *array;
int seive[seive_len];
int prime_prime[seive_len];
int min(int a, int b) {
return a<b?a:b;
}
int max(int a, int b) {
return a>b?a:b;
}
... |
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