XDU 1130 (快速幂)

题目大意：原题链接

解题思路：sum((p+i*d)*C(n,i))=sum(p*C(n,i)+d*i*C(n,i)) ;
又因为i*C(n,i)=n*C(n-1,i-1) 所以sum((p+i*d)*C(n,i))=sum(p*C(n,i)+d*n*C(n-1,i-1));
然后i又是从0到n,C(n,0)+C(n,1)+......+C(n,n)=2^n;
即sum((p+i*d)*C(n,i)=p*2^n+d*n*2^(n-1).

#include<cstdio>
#define N 542
long long Quickpow(long long a,long long b)
{
    long long ans=;
    a=a%N;
    while(b){
        if(b&)
            ans=ans*a%N;
        a=a*a%N;
        b/=;
    }
    return ans;
}
int main()
{
    long long T,p,d,n;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld%lld",&p,&d,&n);
        long long ans=;
        ans+=(p%N*Quickpow(,n)%N)%N;
        ans+=(d%N*n%N*Quickpow(,n-)%N)%N;
        printf("%lld\n",ans%N);
    }
    return ;
}