Codeforces 687C. The Values You Can Make (dp)

题目链接：http://codeforces.com/problemset/problem/687/C

题目大概说给n个各有价值的硬币，要从它们中选出若干个组合成面值k，而要求的是各个方案里这些选出的硬币能组合出来的面值有哪些。

dp[i][j][k]表示到第i个硬币，组成面值为j，包含面值为k的方案数。

注意用滚动数组写。

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 505;
int dp[2][N][N], a[N];

int main()
{
    int n, k;
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", a + i);
    }
    dp[0][0][0] = 1;
    for(int i = 1; i <= n; ++i) {
        for(int j = k; j >= 0; --j) {
            for(int x = j; x >= 0; --x) {
                //memset(dp[i%2], 0, sizeof(dp[i%2])); 
                dp[i%2][j][x] |= dp[(i - 1)%2][j][x]; //滚动数组重复使用
                if(j - a[i] >= x)
                    dp[i%2][j][x] |= dp[(i - 1)%2][j - a[i]][x];
                if(x >= a[i])
                    dp[i%2][j][x] |= dp[(i - 1)%2][j - a[i]][x - a[i]];
            }
        }
    }
    int cnt = 0, ans[505];
    for(int i = 0; i <= k; ++i) {
        if(dp[1][k][i] || dp[0][k][i]) {
            ans[++cnt] = i;
        }
    }
    printf("%d\n", cnt);
    for(int i = 1; i <= cnt; ++i) {
        printf("%d%c", ans[i], i == cnt ? '\n': ' ');
    }
    return 0;
}