Codeforces 577B Modulo Sum 【多重背包】

题目链接：Codeforces 577B Modulo Sum

B. Modulo Sum
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Input
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).

Output
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.

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
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.

题意：给你n个数，问你能否找出若干个数，使得它们之和可以被m整除。

思路：找出余数，最坏情况下，余数最多有10^3个，每个最多有10^3个，这样就变成用这些余数去装m的背包，特判掉余数为0的情况。然后用多重背包优化下，时间复杂度最坏O(2*10^7)。
AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <cmath>#include <stack>#include <map>#include <vector>#define fi first#define se second#define ll o<<1#define rr o<<1|1#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MOD = 1e9 + 7;const int MAXN = 1e6 + 10;const int INF = 1e9 + 10;void add(LL &x, LL y) { x += y; x %= MOD; }int a[MAXN];int num[1001];int val[25*1001];int dp[25*1001][1001];int main(){    int n, m;    while(scanf("%d%d", &n, &m) != EOF) {        for(int i = 0; i < m; i++) {            num[i] = 0;        }        bool flag = false;        for(int i = 0; i < n; i++) {            scanf("%d", &a[i]);            if(a[i] % m == 0) {                flag = true;            }            a[i] %= m; num[a[i]]++;        }        if(flag) {            printf("YES\n");            continue;        }        sort(a, a+n);        int N = unique(a, a+n) - a;        int k = 1;        for(int i = 0; i < N; i++) {            for(int j = 1; j <= num[a[i]]; j <<= 1) {                if((j * a[i]) % m == 0) {                    flag = true;                }                val[k++] = (j * a[i]) % m;                num[a[i]] -= j;            }            if(num[a[i]] > 0) {                if((a[i] * num[a[i]]) % m == 0) {                    flag = true;                }                val[k++] = (num[a[i]] * a[i]) % m;            }        }        if(flag) {            printf("YES\n");            continue;        }        CLR(dp, 0); dp[1][0] = dp[1][val[1]] = 1;        for(int i = 1; i < k-1; i++) {            for(int j = 0; j < m; j++) {                if(dp[i][j]) {                    dp[i+1][j] = 1;                    if((j + val[i+1]) % m == 0) {                        flag = true;                        break;                    }                    dp[i+1][(j+val[i+1])%m] = 1;                }            }            if(flag) break;        }        if(flag) {            printf("YES\n");        }        else {            printf("NO\n");        }    }    return 0;}