lightoj1233_二进制优化

http://lightoj.com/volume_showproblem.php?problem=1233

一维多重背包二进制优化的思考：先看这个问题，100=1+2+4+8+16+32+37，观察可以得出100以内任何一个数都可以由以上7个数选择组合得到，所以对物品数目就不是从0都100遍历，而是0，1，2，4，5，16，32，37遍历，时间大大优化。

In a strange shop there are n types of coins of value A₁, A₂ ... A_n. C₁, C₂, ... C_n denote the number of coins of value A₁, A₂ ... A_n respectively. You have to find the number of different values (from 1 to m), which can be produced using these coins.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 100), m (0 ≤ m ≤ 10⁵). The next line contains 2n integers, denoting A₁, A₂ ... A_n, C₁, C₂ ... C_n (1 ≤ A_i ≤ 10⁵, 1 ≤ C_i ≤ 1000). All A_i will be distinct.

Output

For each case, print the case number and the result.

Sample Input	Output for Sample Input
2 3 10 1 2 4 2 1 1 2 5 1 4 2 1	Case 1: 8 Case 2: 4

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <ctime>
#include <queue>
#include <list>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 1e5 + 5;
int dp[N];
int val[105], c[105];

int main()
{
    int t, n, m;
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca) {
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", val + i);
        }
        for(int i = 1; i <= n; ++i) {
            scanf("%d", c + i);
        }
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for(int i = 1; i <= n; ++i) {
            for(int k = 1; k <= c[i]; k <<= 1) {
                for(int j = m; j >= k*val[i]; --j) {
                    dp[j] |= dp[j - k*val[i]];
                }
                c[i] -= k;
            }
            if(c[i]) {
                for(int j = m; j >= c[i]*val[i]; --j) {
                    dp[j] |= dp[j - c[i]*val[i]];
                }
            }
        }
        int ans = 0;
        for(int i = 1; i <= m; ++i) {
            ans += dp[i];
        }
        printf("Case %d: %d
", ca, ans);
    }
    return 0;
}