题意

eddy走一个长度为\(n\)的环，每次能往前或往后走一步，问走到\(m\)点恰好走完所有点至少一次的概率，前\(i\)个询问的答案要乘起来

分析

  • \(n=1,m=0\),答案为\(1\)
  • \(n>1,m=0\),答案为\(0\)
  • \(n>1,m \ne 0\),答案为\(1/(n-1)\)

Code

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define ll long long
using namespace std;
const int inf=1e9;
const int mod=1e9+7;
const int maxn=1e5+10;
int T;
ll n,m;
ll ksm(ll a,ll b){
	ll ret=1;
	while(b){
		if(b&1) ret=ret*a%mod;
		b>>=1;
		a=a*a%mod;
	}
	return ret;
}
int main(){
	//ios::sync_with_stdio(false);
	//freopen("in","r",stdin);
	scanf("%d",&T);
	ll ans=1;
	while(T--){
		scanf("%lld%lld",&n,&m);
		if(n==1) printf("%lld\n",ans);
		else{
			if(m==0){
				ans=0;
				printf("%lld\n",ans);
			}else{
				printf("%lld\n",(ans*=ksm(n-1,mod-2))%=mod);
			}
		}
	}
	return 0;
}