hud-5776
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Problem Description
Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO
Input
The first line of the input has an integer T (1≤T≤10 ), which represents the number of test cases.
For each test case, there are two lines:
1.The first line contains two positive integers n, m (1≤n≤100000 , 1≤m≤5000 ).
2.The second line contains n positive integers x (1≤x≤100 ) according to the sequence.
For each test case, there are two lines:
1.The first line contains two positive integers n, m (
2.The second line contains n positive integers x (
Output
Output T lines, each line print a YES or NO.
Sample Input
23 31 2 35 76 6 6 6 6
Sample Output
YESNO
Source
BestCoder Round #85
题解:抽屉原理。
当n>m时,有n个前缀和,对m取余就有n个余数,由于n>m,则至少有两个前缀和余数相等,那么相减必之后对m取余必为0.
#include"stdio.h"
#include"cstdio"
#include"algorithm"
#include"string.h"
using namespace std;
const int max_n=5*(1e3)+10;
int A[max_n];
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
memset(A,0,sizeof(A));
int i,n,m;
scanf("%d%d",&n,&m);
if(n>m)
{
int x;
for(i=0;i<n;i++)
scanf("%d",&x);
printf("YES\n");
}
else
{
int x,flage=0;
for(i=1;i<=n;i++)
{
scanf("%d",&x);
A[i]=A[i-1]+x;
}
for(i=0;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
if((A[j]-A[i])%m==0)
{//printf("**%d %d\n",A[j],j);
flage=1;
i=n;
break;
}
}
}
if(flage)
printf("YES\n");
else
printf("NO\n");
}
}
}
}
#include"cstdio"
#include"algorithm"
#include"string.h"
using namespace std;
const int max_n=5*(1e3)+10;
int A[max_n];
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
memset(A,0,sizeof(A));
int i,n,m;
scanf("%d%d",&n,&m);
if(n>m)
{
int x;
for(i=0;i<n;i++)
scanf("%d",&x);
printf("YES\n");
}
else
{
int x,flage=0;
for(i=1;i<=n;i++)
{
scanf("%d",&x);
A[i]=A[i-1]+x;
}
for(i=0;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
if((A[j]-A[i])%m==0)
{//printf("**%d %d\n",A[j],j);
flage=1;
i=n;
break;
}
}
}
if(flage)
printf("YES\n");
else
printf("NO\n");
}
}
}
}
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