题意：给你n个数，从中随机取几个数，使得这几个数的和是m的倍数，如果存在这样一组数，输出yes，否则输出no
思路：首先，当n>m时，肯定有符合条件的，由同余定理可以得到，将这前n项和对m取余的结果用一个数组表示，则因为n大于m，必然存在两个相等的数s1,和s2，则这两个数中间的那些数字之和就能整除m。其次，对于n<m的情况，用一个数组dp[i][j]标记这个和有没有出现过，如果没有，那么就标记为1，dp[i][j]表示i个数里，和为j的这个数字有没有出现过，最后只需要判断dp[n][m]是不是等于1就行
#include<bits\stdc++.h>
using namespace std;
const int maxn = 1e6+6;
int a[maxn];
int dp[1005][1005];
int main()
{
int n,m;
int flag = 0;
cin >> n >> m;
for(int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]%=m;
if(a[i]==0)
flag=1;
}
if(n>m || flag)
{
puts("YES");
return 0;
}
memset(dp,0,sizeof(dp));
for(int i = 0;i<=n;i++)
dp[i][0]=1;
for(int i = 1;i<=n;i++)
for(int j = 0;j<=m;j++)
dp[i][j]=(dp[i-1][(j+a[i])%m] || dp[i-1][j]);
if(dp[n][m])
puts("YES");
else
puts("NO");
}
B. Modulo Sum
time limit per test
memory limit per test
input
output
a1, a2, ..., an, and a number m.
aij such that the sum of numbers in this subsequence is divisible by m.
Input
n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
Output
YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.
Examples
input
3 5 1 2 3
output
YES
input
1 6 5
output
NO
input
4 6 3 1 1 3
output
YES
input
6 6 5 5 5 5 5 5
output
YES
Note
2 and 3, the sum of which is divisible by 5.
5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
3
In the fourth sample test you can take the whole subsequence.
