Hasan and Bahosain want to buy a new video game, they want to share the expenses. Hasan has a set of N coins and Bahosain has a set of M coins. The video game costs W JDs. Find the number of ways in which they can pay exactly W JDs such that the difference between what each of them payed doesn’t exceed K.
In other words, find the number of ways in which Hasan can choose a subset of sum S1and Bahosain can choose a subset of sum S2 such that S1 + S2 = W and |S1 - S2| ≤ K.
Input
The first line of input contains a single integer T, the number of test cases.
The first line of each test case contains four integers N, M, K and W (1 ≤ N, M ≤ 150) (0 ≤ K ≤ W) (1 ≤ W ≤ 15000), the number of coins Hasan has, the number of coins Bahosain has, the maximum difference between what each of them will pay, and the cost of the video game, respectively.
The second line contains N space-separated integers, each integer represents the value of one of Hasan’s coins.
The third line contains M space-separated integers, representing the values of Bahosain’s coins.
The values of the coins are between 1 and 100 (inclusive).
Output
For each test case, print the number of ways modulo 109 + 7 on a single line.
Example
2
4 3 5 18
2 3 4 1
10 5 5
2 1 20 20
10 30
50
2
0
对A和B的硬币分别dp，f[i]表示拼成i元的方案数，for i=1 to n for j=15000 down to 0 f(j+a(i))+=f(j)
最后枚举一下差在K以内，且i+j=W的f(i)和g(j)即可。
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
#define MOD 1000000007ll
int T,n,m,K,W;
int a[160],b[160];
int Abs(int x)
{
return x<0 ? (-x) : x;
}
ll f[15010],g[15010];
int main()
{
scanf("%d",&T);
for(;T;--T)
{
scanf("%d%d%d%d",&n,&m,&K,&W);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
for(int i=1;i<=m;++i)
scanf("%d",&b[i]);
memset(f,0,sizeof(f));
memset(g,0,sizeof(g));
f[0]=1;
for(int i=1;i<=n;++i)
for(int j=15000;j>=0;--j)
if(a[i]+j<=15000)
f[j+a[i]]=(f[j+a[i]]+f[j])%MOD;
g[0]=1;
for(int i=1;i<=m;++i)
for(int j=15000;j>=0;--j)
if(b[i]+j<=15000)
g[j+b[i]]=(g[j+b[i]]+g[j])%MOD;
ll ans=0;
for(int i=0;i<=W;++i)
if(Abs(W-i-i)<=K)
ans=(ans+f[i]*g[W-i]%MOD)%MOD;
cout<<ans<<endl;
}
return 0;
}