原题地址:http://acm.fzu.edu.cn/problem.php?pid=2301

题意:有n次攻击回合，每个回合的攻击有一点伤害，但是可能攻击到敌方无限血的英雄，也可能攻击到敌方的一个血量为m的小兵，问在n次攻击回合内把敌方小兵杀死的概率为多少？答案对1e9+7取模。

组合数数学公式,首先总共有$2^n$的组合方法,然后要造成m点伤害及以上,所以就是$C^{m}_{n}+C^{m+1}_{n}+C^{m+2}_{n}+......+C^{n}_{n}$.然后答案就是两者除一下.因为$n$只有1000,预处理就行了.

#include <cmath>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <map>
#include <cctype>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
ll Pow(ll base, ll n) {
    ll res = 1;
    while (n) {
        if (n & 1) res = res * base % mod;
        n >>= 1;
        base = base * base % mod;
    }
    return res;
}

ll C[1005][1005];
//C[n][m]就是C(n,m)
void init(int N) {
    for (int i = 0; i < N; i ++) {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; j ++) {
            C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
            C[i][j] %= mod;
        }
    }
}

int main() {
    init(1005);
    int t;
    scanf("%d", &t);
    while (t--) {
        int n, m;
        scanf("%d%d", &n, &m);
        ll ans = 0;
        for (int i = m; i <= n; i++) ans = (ans + C[n][i]) % mod;
        ll sum = Pow(2, n);
        printf("%lld\n", ans * Pow(sum, mod - 2) % mod);
    }
    return 0;
}