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最新推荐文章于 2020-09-15 21:06:08 发布

fire_lch316

最新推荐文章于 2020-09-15 21:06:08 发布

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本文链接：https://blog.youkuaiyun.com/fire_lch316/article/details/77524114

本文探讨了一道算法题目，即判断给定序列是否存在连续子序列使得其和能被特定整数m整除的问题。文章提供了详细的解决方案，利用抽屉原理进行证明，并通过示例输入输出展示了解决方案的有效性。

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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.

Output

Output T lines, each line print a YES or NO.

Sample Input

Sample Output

  
   YES
NO

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");
}
}
}
}