#include <iostream> #include <string> #include <algorithm> #include <vector> using namespace std; typedef long long ll; //(a+b)%c=a%c+b%c; 将大数才成一位一位逐个乘10相加 //(a*b)%c=(a%c*b%c)%c ll func(string a,ll c){     ll sum=0;     for(i...

先把大数取模，然后就变成了快速幂的简单问题， https://blog.csdn.net/csyifanZhang/article/details/105623224 ↑求模运算的性质总结   // s%k int Mod(string s, int k) { int res = 0, mod = 0; for (int i = 0; i < s.size(); i++) { res = mod * 10 + s[i] - '0'; s[i] = res / k; mod = res % k; } ret...

import java.io.*; import java.math.BigInteger; import java.util.*; public class Main {             public static void main(String[] args) {         Scanner cin = new Scanner(System.in);      &...