题目链接：Problem - D - Codeforces
题目大意：
一共n轮比赛，有\(2^n\)个参赛者，第\(i\)轮有\(2^{n - i}\) 场比赛，Madoka能安排第一局的比赛，她想让最后的赢家编号更小，主办方最多有k次操作，能修改任意每一场比赛的获胜情况，可以让最终赢家编号更
大，求Madoka在主办方任意修改之后可能获得的胜利者的最小编号
比如上面就是整个比赛，Madoka安排的第一局的顺序是 3 - 1 4 - 2，最终的结果是1获胜，但是第二个图是主办方修改之后的，导致3获胜了
算法过程
首先，我们可以让这个二叉树的左节点是获胜的那一方，右节点是输的那一方，然后我们看k次操作最多把第一局的那几个节点变成最终赢家，那么这就得满足从二叉树的根节点，到这个节点输的边得少于等于k，那么我求所有输的边有1个，输的边有2个......输的边有k个就一定能把这个点变成最终赢家，那么，主办方最多能变这么多，Madoka就能按照这个来安排开始的顺序，最小的编号就是能除去这么多一定能变成最终节点的节点，我们只需要求所有小于k个输边的数量就行，要用到组合数学，因为每个节点到最终节点的路上一共n条边，求\(C^1_n\) + \(C^2_n\) + ......\(C^k_n\) 就是最终结果
代码
// Problem: D. Madoka and The Corruption Scheme
// Contest: Codeforces - Codeforces Round 818 (Div. 2)
// URL: https://codeforces.com/contest/1717/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//by codeforcer ——
// ____ _ _ _ _ _ _ ____ _
// / ___|| | | || | | || | | | |___ \ | |
//| | | |_| || |_| || |_| | __) | | |
//| | | _ || _ || _ | |__ < | |
//| |___ | | | || | | || | | | ___) | | |
// \____||_| |_||_| |_||_| |_| |____ / |_|#include<bits/stdc++.h>
using namespace std;typedef int E;
typedef long long LL;
typedef pair<int, int> PII;
typedef tuple<int, int, int> PIII;
typedef tuple<LL, LL, LL> PLLL;
typedef pair<long long, long long> PLL;
typedef unsigned long long ULL;
#define int long long
#define lowbit(x) (x & (-x))
#define mod(x) (x % MOD + MOD) % MOD
#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)
#define endl '\n'
#define vec vector
#define pb push_back
#define pob pop_back
#define fir first
#define sec second
#define INT 0x3f3f3f3f
#define LLONG 0x3f3f3f3f3f3f3f3fLL
#define lower lower_bound
#define upper upper_bound
#define umap unordered_map
#define uset unordered_set
#define maxheap priority_queue<E, vector<E>, less<E>>
#define minheap priority_queue<E, vector<E>, greater<E>>#define prvec(a) \for (int i = 0; i < (a).size(); i++) { \cout << (a)[i] << " "; \} \cout << endl;#define debug0(a) \cout << #a << ": "; \for (int k = 0; k < (a).size(); k++) { \cout << (a)[k] << " "; \} \cout << endl; #define debug1(a) \cout << #a << ": "; \for (int k = 1; k <= (a).size(); k++) { \cout << (a)[k] << " "; \} \cout << endl; LL gcd(LL a, LL b) { return (b) ? gcd(b, a % b) : a; }
LL exgcd(LL a, LL b, LL &x, LL &y) {if (b == 0) { x = 1, y = 0; return a; }LL gcd = exgcd(b, a % b, y, x);y -= a / b * x;return gcd;}
LL qmi(LL a, LL b, LL mod) {LL res = 1;while (b) {if (b & 1) res = res * a % mod;a = a * a % mod;b >>= 1;}return res;}const int N = 100010,MOD = 1e9 + 7;const int mod = 1e9 + 7;
int fact[N], infact[N];int pmod(int a,int b){int res=1;while(b){if(b&1) res=res*a%mod;b>>=1;a=a*a%mod;}return res;
}void init()
{fact[0] = 1, infact[0] = 1;for (int i = 1; i <= N; i++){fact[i] = fact[i - 1] * i % mod;infact[i] = infact[i - 1] * pmod(i,mod - 2) % mod;}
}int C(int a,int b)
{return fact[a] * infact[a - b] % mod * infact[b] % mod;
}
void solve() {int n,k;cin>>n>>k;init();int ans = 0;for(int i = 0;i <= min(k,n);i ++){ans = (ans + C(n,i)) % mod;}cout<<ans<<endl;}signed main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);//i//nt n;cin>>n;while(n --)solve();return 0;
}
