ZOJ 3924 Musical Notes

Description

Besides his skills of playing Hearthstone, Bob is also insterested in learning piano in order to make himself more charming, of course, to Alice.

Now, Bob had transcribed a musical score for himself to practice. Unfortunately, he only copied the numbered musical notation, which use numbers 1 to 7 representing the musical notes(sung as do, re, mi...) and he forgot to mark the musical notes, which means he did't know the lengths of those notes. Bob remembered that the composition contains m bar and k

Knowing  m and 

k, he can calculate the whole length of those musical notes, 

l beats, which equals 

4*m-k beats. Then he counted out 

n, the number of notes represented by number 1 to 7, with which he may find out how the composition is formed, namely finding out the length of each note. He thought there might be too many cases to form the composition, so he want you to find out the number of possible ways to form the composition.

Input

The first line of input contains an integer T (T ≤ 20) . T is the number of the cases. In the next T lines, there are three integer m, k, n, representing the number of bars, the number of quarter rests, and the number of musical notes.(1 ≤ m ≤ 10000, 0 ≤ k ≤ 3, m ≤ n ≤ m+13)

Output

The output contains T

Sample Input

12 0 5      

Sample Output

75      

Hint

The number of whole, half, quarter, eighth, sixteenth notes can be:

0, 3, 2, 0, 0. In this situation we may find 10 ways to form the composition.

1, 0, 4, 0, 0. In this situation we may find 5 ways.

1, 1, 1, 2, 0. In this situation we may find 60 ways.

简化一下，题意就是说有5种节拍分别是4，2，1，0.5，0.25然后总共有n个，这些节拍的和是4*m-k

并且呢，n的范围是[m,m+13]，问的是这样的组合方案有多少种，于是直接dfs找出全部的可能乘上组合数

因为直接求组合数比较麻烦，我是直接用全排列除掉各自的排列数得到答案的，因为要取余，所以事先算好

#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e4+10;
int a[5]={16,8,4,2,1};
int b[5];
int T,n,k,m;
LL ans,f[maxn],c[maxn];

LL inv(LL x, LL m)  
{  
    if (x == 1) return x;  
    return inv(m % x, m)*(m - m / x) % m;  
}  

void init()
{
    c[0]=f[0]=1;
    for (int i=1;i<maxn;i++) 
    {
        c[i]=c[i-1]*i%mod;
        f[i]=inv(c[i],mod);
    }
}

void dfs(int x,int y,int z)
{
    if (x*a[z]<y) return;
    if (z==4)
    {
        if (x==y)
        {
            b[4]=x;
            LL temp=c[n];
            for (int i=0;i<5;i++) temp=temp*f[b[i]]%mod;
            (ans+=temp)%=mod;
        }
        return;
    }
    for (int i=0;i<=x;i++)
    {
        b[z]=i;
        dfs(x-i,y-a[z]*i,z+1);
    }
}

int main()
{
    init();
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d%d",&m,&k,&n);
        ans=0;
        dfs(n,16*m-4*k,0);
        printf("%lld\n",ans);
    }
    return 0;
}