Algorithm
You are given a sequence of numbers a1, a2, ..., an, and a number m.
Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.
The first line contains two numbers, 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.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.
3 5
1 2 3
YES
1 6
5
NO
4 6
3 1 1 3
YES
6 6
5 5 5 5 5 5
YES
In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
In the third sample test you need to choose two numbers 3 on the ends.
In the fourth sample test you can take the whole subsequence.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <stdio.h>
#include <memory.h>
using namespace std;
int const N = 1e6 + 1;
int n, m, a[N], dp[1001][1001][2];
bool rec(int i, int sum, bool take) {
if(sum == 0 && take)
return true;
if(i == n)
return false;
int &ret = dp[i][sum][take];
if(ret != -1) return ret;
ret = false;
return ret = (rec(i + 1, (sum + a[i]) % m, true) || rec(i + 1, sum % m, take));
}
int main() {
scanf("%d %d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%d", a + i);
if(n > m) {
puts("YES");
return 0;
}
memset(dp, -1, sizeof dp);
if(rec(0, 0, 0))
puts("YES");
else
puts("NO");
return 0;
}Input
1 2 3
Output
Demonstration
Codeforces Solution-Modulo Sum-Solution in C, C++, Java, Python