2018 ICPC焦作网络赛 K Transport Ship（多重背包二进制优化）

There are N different kinds of transport ships on the port. The i^th kind of ship can carry the weight of V[i]V[i] and the number of the i^th kind of ship is 2^c[i]-12^{C[i]} - 12. How many different schemes there are if you want to use these ships to transport cargo with a total weight of S?

It is required that each ship must be full-filled. Two schemes are considered to be the same if they use the same kinds of ships and the same number for each kind.

Input

The first line contains an integer T(1≤T≤20), which is the number of test cases.

For each test case:

The first line contains two integers: N(1≤N≤20), Q(1≤Q≤10000), representing the number of kinds of ships and the number of queries.

For the next N lines, each line contains two integers: V[i](1≤V[i]≤20), C[i](1≤C[i]≤20), representing the weight the i^th kind of ship can carry, and the number of the i^th kind of ship is 2^c[i]-1.

For the next Q lines, each line contains a single integer: S(1≤S≤10000), representing the queried weight.

Output

For each query, output one line containing a single integer which represents the number of schemes for arranging ships. Since the answer may be very large, output the answer modulo $1000000007$.

样例输入

1
1 2
2 1
1
2

样例输出

0
1

题意：

有N种船只，每种船只的载货量为V[i]（以下代码用w[i]表示），每种船只的数量为2^c[i]-1。接下来有Q次询问，每次问有多少种载货方式可以填满容量S。

思路：

如果用裸的01背包的话时间复杂度是O(N*2^c[i]*10000)，显然会超时，但是我们可以把每一种船合并，比如船只有2ⁿ-1艘的话，就拆成2⁰+2¹+2²+...+2^n-1，1~2ⁿ-1之中的任意一个数都可以由拆分出来的数组成，将所有合并后的结果进行一次多重背包即可。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e4;
const int mod=1e9+7;
typedef long long ll;
int w[25],c[25];
ll dp[MAX+5];
int main()
{
    int n,i,T,q,s,j,k;
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--)
    {
        cin>>n>>q;
        for(i=1;i<=n;i++)
            cin>>w[i]>>c[i];
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(i=1;i<=n;i++)       //共n种船
        {
            int t=1;
            for(j=1;j<=c[i];j++)//每种船有2^c[i]-1只
            {
                for(k=MAX;k>=t*w[i];k--)
                    dp[k]=(dp[k]+dp[k-t*w[i]])%mod;
                t<<=1;
            }
        }
        while(q--)
        {
            cin>>s;
            cout<<dp[s]<<endl;
        }
    }
    return 0;
}