焦作网络赛K-Transport Ship【dp】

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

 

题意：

有n种船 每种有2^c[i] - 1艘 每艘载重v[i]【必须完全装满】

现给出q次查询s 输出装载s的安排船的方案数

思路：

形如多重背包问题 用二进制的思想就行优化

dp[w]表示载重w时的方案数 他可由dp[w - v[i] * k]得到 其中k是系数

按照二进制的思想 每次k*=2 最多20次 并且判断和c[i]的关系

要注意因为是2^c[i] -1因此等号不能取

 

// ConsoleApplication3.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//

//#include "pch.h"
#include <iostream>
#include<algorithm>
#include<stdio.h>
#include<set>
#include<cmath>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<stack>

#define inf 0x3f3f3f3f

using namespace std;
typedef long long LL;

const int maxn = 25;
const int maxs = 10005;
const LL mod = 1000000007;
int t, n, q, s;
int v[maxn], c[maxn];
LL dp[maxs];

int main()
{
    cin >> t;
    while (t--) {
        scanf("%d%d", &n, &q);
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &v[i], &c[i]);
        }

        memset(dp, 0, sizeof(dp));
        dp[0] = 1;

        int k = 1;
        for (int j = 0; j < 20; j++) {
            for (int i = 0; i < n; i++) {
                if (j >= c[i]) {
                    continue;
                }
                for (int w = maxs; w >= v[i] * k; w--) {
                    dp[w] += dp[w - k * v[i]];
                    dp[w] %= mod;
                }
            }
            k *= 2;
        }

        while (q--) {
            scanf("%d", &s);
            printf("%lld\n", dp[s]);
        }
    }
}