题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5776
题目让你求是否有区间的和是m的倍数。
预处理前缀和，一旦有两个数模m的值相同，说明中间一部分连续子列可以组成m的倍数。
1 //#pragma comment(linker, "/STACK:102400000, 102400000")
2 #include <algorithm>
3 #include <iostream>
4 #include <cstdlib>
5 #include <cstring>
6 #include <cstdio>
7 #include <vector>
8 #include <cmath>
9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long LL;
15 typedef pair <int, int> P;
16 const int N = 1e5 + 5;
17 int a[N], cnt[N];
18
19 int main()
20 {
21 int t, n, k;
22 scanf("%d", &t);
23 while(t--) {
24 scanf("%d %d", &n, &k);
25 memset(cnt, 0, sizeof(cnt));
26 bool ok = false;
27 for(int i = 1 ; i <= n ; ++i) {
28 scanf("%d", a + i);
29 a[i] = (a[i] + a[i - 1]) % k;
30 cnt[a[i]]++;
31 if(cnt[a[i]] > 1 || !a[i])
32 ok = true;
33 }
34 if(ok)
35 printf("YES\n");
36 else
37 printf("NO\n");
38 }
39 return 0;
40 }
sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 264 Accepted Submission(s): 127
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
Source