Chosen by god【组合数打表】

Chosen by god

 题目链接（点击）

Everyone knows there is a computer game names "hearth stone", recently xzz likes this game very much. If you want to win, you need both capacity and good luck.

There is a spell card names "Arcane Missiles", which can deal 3 damages randomly split among all enemies.

xzz have a "Arcane Missiles Plus" can deal n damage randomly split among all enemies. The enemy battle field have only two characters: the enemy hero and enemy minion.

Now we assume the enemy hero have infinite health, the enemy minion has m health.

xzz wants to know the probability of the "Arcane Missiles Plus" to kill the enemy minion (the minion has not more than 0 health after "Arcane Missiles Plus"), can you help him?

Input

The first line of the input contains an integer T, the number of test cases. T test cases follow. (1 ≤ T ≤ 100000)

Each test case consists of a single line containing two integer n and m. (0 ≤ m ≤ n ≤ 1000)

Output

For each test case, because the probability can be written as x / y, please output x * y^-1 mod 1000000007. .(y * y^-1 ≡ 1 mod 10^9 + 7)

Sample Input

2
3 3
2 1

Sample Output

125000001
750000006

题意：

给出T组n和m n代表你有n发子弹 敌人有两个 一个血量无穷大 一个为m 每次击中敌人都会掉1滴血 问把血量为m的敌人打死的几率有多大 每次开枪随机打中其中一个敌人

思路：

概率问题 先找所有情况 2^n 击中的概率 c(n,m)+c(n,m+1)+……+c（n,n）

所以p=c(n,m)+c(n,m+1)+……+c（n,n)/2^n

看到组合数求和就有点怕了 想到了组合数打表 之前只是听到过

杨辉三角打表组合数、组合数求和：

仔细观察杨辉三角 有如下规律：

第一行表示c（0,0）n=0

第二行：c（1,0）c（1,1）n=1

以后可以递推，并且每个数的值是其肩膀上两个值的和 根据这个可以打表 代码如下：

void getc()
{
    c[0][0]=1,sum[0][0]=1;
    for(int i=1;i<=1001;i++){
        for(int j=0;j<=i;j++){
            if(j==0||j==i){
                c[i][j]=1;
            }
            else{
                c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
            }
            if(j==0){
                sum[i][j]=1;
            }
            else{
                sum[i][j]=(sum[i][j-1]+c[i][j])%mod;
            }
        }
    }
}

复杂度高了点 但用这个题还好没T

但如果不去先sum求和打好表 就不好说了

AC代码：

下面代码里面也用到了取余的公式 特别是减法 WA了好多次：https://mp.csdn.net/postedit/89529166

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string>
#include<string.h>
#include<map>
using namespace std;
typedef long long LL;
const int MAX=1e6;
const LL mod=1e9+7;
const LL INF=0x3f3f3f3f;
LL qpow(LL m,LL q)
{
    LL ans=1;
    while(q){
        if(q%2){
            ans=ans*m%mod;
        }
        m=m*m%mod;
        q/=2;
    }
    return ans%mod;
}
LL getinv(LL x,LL y)
{
    LL ans=qpow(x,y-2)%mod;
    return ans;
}
LL c[1005][1005],sum[1005][1005];
void getc()
{
    c[0][0]=1,sum[0][0]=1;
    for(int i=1;i<=1001;i++){
        for(int j=0;j<=i;j++){
            if(j==0||j==i){
                c[i][j]=1;
            }
            else{
                c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
            }
            if(j==0){
                sum[i][j]=1;
            }
            else{
                sum[i][j]=(sum[i][j-1]+c[i][j])%mod;
            }
        }
    }
}
int main()
{
    getc();
    LL T,n,m;
    cin>>T;
    while(T--){
        LL num=0,sum1=0;
        cin>>n>>m;
        num=(sum[n][n]-sum[n][m-1]+mod)%mod;
        sum1=getinv(qpow(2,n),mod);
        sum1=(sum1*num)%mod;
        cout<<sum1<<"
";
    }
    return 0;
}