题意：6个物品，价值为1 2 3 4 5 6，每次告诉你每个物品有多少个，问你是否能选出一些物品，使得这些物品的价值为所有物品总价值的一半

思路：直接多重背包，加个二进制优化防超时

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e6 + 10;
int kase = 0, a[10], b[maxn];
bool dp[maxn];
int main() {
    while (~scanf("%d %d %d %d %d %d", &a[1],&a[2],&a[3],&a[4],&a[5],&a[6])) {
        if (a[1]+a[2]+a[3]+a[4]+a[5]+a[6] == 0) break;
        int sum = 0;
        for (int i = 1; i <= 6; i++) sum += a[i]*i;
        printf("Collection #%d:\n", ++kase);
        if (sum & 1) {
            printf("Can't be divided.\n\n"); continue;
        }
        int cnt = 0;
        for (int i = 1; i <= 6; i++) {
            int j = 0;
            while (a[i] >= (1<<j)) {
                b[++cnt] = (1<<j)*i;
                a[i] -= (1<<j);
                j++;
            }
            if (a[i] > 0) {
                b[++cnt] = a[i]*i;
            }
        }
        int mid = sum / 2;
        for (int i = 0; i <= mid; i++) dp[i] = false;
        dp[0] = true;
        for (int i = 1; i <= cnt; i++) {
            for (int j = mid; j >= b[i]; j--) {
                if (dp[j-b[i]]) dp[j] = true;
            }
        }
        if (dp[mid]) printf("Can be divided.\n");
        else printf("Can't be divided.\n");
        printf("\n");
    }
    return 0;
}