Hdu5776 字串的和[鸽巢原理]

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题目连接

http://acm.hust.edu.cn/vjudge/problem/440424

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 ( ), 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 ( , ).
2.The second line contains n positive integers x ( ) according to the sequence.

Output

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

Sample Input

2
3 3
1 2 3
5 7
6 6 6 6 6

Sample Output

YES
NO

题意

给出n个数字构成的串，问其中是否存在一个连续字串的和是m的倍数。

题解

对前缀和做取m余运算，可证若存在两个前缀和取余结果相同，则由这两个前缀和构成的字串的和是m的倍数[鸽巢原理可证]。对n为1时要特判一下。

代码

#include<iostream>#include<queue>  #include<cstdio>  #include<cstring>  #include<string>  #include<cmath>  #include<string> #include<cctype>  #include<algorithm>  #include<vector>using namespace std;const int maxn=100005;int main(){    int t;    scanf("%d",&t);    while(t--)    {   int a[maxn];        int b[maxn];        int s[maxn];        memset(b,0,sizeof(b));        int n,m,sum=0;int ok=0;        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        s[0]=a[0];        for(int i=1;i<n;i++){a[i]+=a[i-1];s[i]=a[i];}        for(int i=0;i<n;i++){s[i]%=m;b[s[i]]++;}        for(int i=0;i<=m;i++)        {            if(b[i]>1){ok=1;break;}            if(b[0]==1){ok=1;break;}        }        if(ok==1)printf("YES\n");        else printf("NO\n");    }return 0;}

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