ACM-ICPC 2018 焦作赛区网络预赛 K.Transport Ship（01背包）

There are NN different kinds of transport ships on the port. The i^{th}ith kind of ship can carry the weight of V[i]V[i] and the number of the i^{th}ith kind of ship is 2^{C[i]} - 12C[i]−1. How many different schemes there are if you want to use these ships to transport cargo with a total weight of SS?

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 \le T \le 20)T(1≤T≤20), which is the number of test cases.

For each test case:

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

For the next NN lines, each line contains two integers: V[i](1 \le V[i] \le 20), C[i](1 \le C[i] \le 20)V[i](1≤V[i]≤20),C[i](1≤C[i]≤20), representing the weight the i^{th}ith kind of ship can carry, and the number of the i^{th}ith kind of ship is 2^{C[i]} - 12C[i]−1.

For the next QQ lines, each line contains a single integer: S(1 \le S \le 10000)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 10000000071000000007.

样例输入复制

1
1 2
2 1
1
2

样例输出复制

0
1

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

题目链接：https://nanti.jisuanke.com/t/31720

题意：有N种船只，每种船只的载货量为v[i]，每种船只有2^c[i]-1种，有q次询问，每次问有多少种载货方式填满容量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)，显然是会超时的，但是我们可以把每一种船只合并，比如船只有2^x-1艘的话，就拆成2^0+2^1+2^2+...+2^(x-1)，1~2^x-1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01背包即可。

#include
using namespace std;
const int mod=1e9+7;
const int maxn=1e4+5;
typedef long long ll;
ll dp[maxn];
int main()
{
    int t,n,q,v[25],c[25],s;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&q);
        for(int i=1;i<=n;i++)
            scanf("%d%d",&v[i],&c[i]);
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        int V;
        for(int i=1;i<=n;i++)
        {
            for(int k=0;k=V;j--)
                {
                    dp[j]+=dp[j-V];
                    dp[j]%=mod;
                }
            }
        }
        for(int i=1;i<=q;i++)
        {
            scanf("%d",&s);
            printf("%lld\n",dp[s]);
        }
    }
}