hdu 5776 抽屉定理

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2547 Accepted Submission(s): 973

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

Sample Output
YES
NO

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=100000+10;int a[maxn];int b[5010];int main() {    int t;    cin>>t;    while(t--) {        memset(b,0,sizeof(b));        int n,m;        cin>>n>>m;        cin>>a[0];        a[0]=a[0]%m;        b[b[0]]++;        int flag=0;        for(int i=1;i<n;i++) {            cin>>a[i];            a[i]=(a[i]+a[i-1])%m;            b[a[i]]++;            if(a[i]==0||b[a[i]]>=2) {                flag=1;            }        }        if(flag) cout<<"YES"<<endl;        else cout<<"NO"<<endl;    }    return 0;}

题解：首先已知：若 x % m = b 且 y % m = b，那么x可以写成x = a1 * m + b，y可以写成y = a2 * m + b，(y - x) % m = ((a2 - a1) * m) % m = 0
当模m后，余数在【0，m)之间，相当于共有m个抽屉，序号从0~m-1,放入m+1个物体，其中必定有两个物体在同一个抽屉之中，利用上述定理得，该序列能整除m

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