求一个数n次方后的末尾数（数论/快速幂）问题描述解题思路代码实现运行结果参考

#include<iostream>
using namespace std;
int main(){
    int m,n,last;

    while(cin>>m>>n){
        last=m%10;
        if(last==0||last==1||last==5||last==6){
            cout<<last<<endl;
        }else if(last==4){
            if(n%2==1){
                cout<<4<<endl;
            }
            if(n%2==0){
                cout<<6<<endl;
            }
        }else if(last==9){
            if(n%2==1){
                cout<<9<<endl;
            }
            if(n%2==0){
                cout<<1<<endl;
            }
        }else if(last==2){
            if(n%4==1){
                cout<<2<<endl;
            }
            if(n%4==2){
                cout<<4<<endl;
            }
            if(n%4==3){
                cout<<8<<endl;
            }
            if(n%4==0){
                cout<<6<<endl;
            }
        }else if(last==3){
            if(n%4==1){
                cout<<3<<endl;
            }
            if(n%4==2){
                cout<<9<<endl;
            }
            if(n%4==3){
                cout<<7<<endl;
            }
            if(n%4==0){
                cout<<1<<endl;
            }
        }else if(last==7){
            if(n%4==1){
                cout<<7<<endl;
            }
            if(n%4==2){
                cout<<9<<endl;
            }
            if(n%4==3){
                cout<<3<<endl;
            }
            if(n%4==0){
                cout<<1<<endl;
            }
        }else if(last==8){
            if(n%4==1){
                cout<<8<<endl;
            }
            if(n%4==2){
                cout<<4<<endl;
            }
            if(n%4==3){
                cout<<2<<endl;
            }
            if(n%4==0){
                cout<<6<<endl;
            }
        }
    }
    return 0;
}

#include <stdio.h>
#include <math.h>

int main() {
    int a[10] = {1, 1, 4, 4, 2, 1, 1, 4, 4, 2};
    int n, num, rmd, ans; // rmd = rightmost digit
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &num);
        rmd = num % 10;
        ans = (int) pow(rmd, num % a[rmd] ? num % a[rmd] : a[rmd]);
        printf("%d\n", ans % 10);
    }
}

#include<iostream>//1097
#include<algorithm>
using namespace std;
int main()
{    int a,b,c[4];
  while(cin>>a>>b)
  {
      a=a%10;
      c[0]=a;//一次方的末尾数
     c[1]=(c[0]*a)%10;//二次方的末尾数
     c[2]=(c[1]*a)%10;//三次方的末尾数
     c[3]=(c[2]*a)%10;//四次方的末尾数
     if(b%4==1)
         cout<<c[0]<<endl;
     if(b%4==2)
         cout<<c[1]<<endl;
     if(b%4==3)
         cout<<c[2]<<endl;
     if(b%4==0)
         cout<<c[3]<<endl;
  }
  return 0;
}