2017-2018 ACM-ICPC Asia East Continent League Final (ECL-Final 2017)

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A

I'am crazy now.
I was wa for many times for a little problem
ans will be litter than sum sometime and I forget make it a positive integer

ans = (ans - sum + mod) % mod;

This is a easy problem that we should get the answer
\(\sum_{i=k}^{n}C(i,n)\)
\(=C_{n}^{k}+ \cdots + C_{n}^{n} \\= (C_{n}^{0}+ C_{n}^{1} + \cdots + C_{n}^{n}) - (C_{n}^{0}+ C_{n}^{1} + \cdots + C_{n}^{k-1})\)

as we know that
\[2^n = (1+1)^{n} = C_{n}^{0} \times 1^{0} \times 1^{n} + C_{n}^{1} \times 1^{1} \times 1^{n-1} + \cdots + C_{n}^{n} \times 1^{n} \times 1^{0} = C_{n}^{0} + C_{n}^{1} + \cdots + C_{n}^{n}\]

so the answer is
\(2^{n} - (C_{n}^{0}+ C_{n}^{1} + \cdots + C_{n}^{k-1})\)

\(C_{n}^{k}÷ C_{n}^{k-1}= \frac{n-k+1}{k}\)

Do't forget the inv() in fraction

#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int mod = 1000000007;
ll pow(ll a,ll b,ll p){
    ll ans = 1;
    a = a % mod;
    while(b){
        if(b&1)ans = ans * a % p;
        b >>= 1;
        a = a * a % p;
    }
    return ans;
}
ll inv(ll a,ll p){
    return pow(a,p-2,p);
}
void getans(ll n,ll k){
    ll ans = pow(2,n,mod)-1;
    ll sum = 1;
    for(ll i=1;i<k;i++){
        sum = ( (sum * (n-i+1) % mod) * inv(i,mod)) % mod;
        ans = (ans - sum + mod) % mod;
    }
    cout<<ans<<endl;
}
int main(){
    ll t;
    cin>>t;
    for(ll i=1;i<=t;i++){
        ll n,k;
        cin>>n>>k;
        printf("Case #%lld: ",i);
        getans(n,k);
    }
    return 0;
}

标签：League,ll,times,sum,cdots,ans,2017,Final,mod
来源： https://www.cnblogs.com/Emcikem/p/12007182.html