#include<bits/stdc++.h>usingnamespacestd;longlongm,n,ans=1,a[5000005];constlonglongmod=100000007;longlongf(longlongx){longlongw=mod-2,s=1,b=x;while(w!=0){if(w&1==1)s=s*b%mod;b=b*b%mod;w>&g

（1）ABmodC.(1<=A,B<2^62,1<=C<=10^9)http://acm.bit.edu.cn/mod/programming/view.php?a=530快速幂----二分#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>usingnamespacestd;longlongquickpow(

这其实是一道洛谷模板题，题目是5435对预处理的讲解可以看看这个博客（代码看自己的，见下）voidgetprime(){ for(inti=0;i<=2;i++)fac[1][i]=1; for(inti=2;i<=N-10;i++) { if(!v[i]) { v[i]=i; prime[++tot]=i; fac[i][0]=fac[i][1]=1; fac[i][2]=i; } fo

#include<iostream>#include<iomanip>#include<ctime>#include<cstdlib>usingnamespacestd;intmain(){  inta,b,gcd;  cin>>a>>b;  gcd=a>b?b:a;  while(gcd>1&&(a%gcd!=0||b%gcd!

L1-096谁管谁叫爹#include<bits/stdc++.h>usingnamespacestd;intdigit_sum(intnum){intsum=0;while(num){sum+=(num%10);num/=10;}returnsum;}intmain(){intN;inti,j;intA,B;cin>>N;for(i=0;i<N;