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 1Sample 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
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