思路1
由于20^22^太大,用double来存储,即(5+20^22^ % 7) % 7即可
java
int num = 5;
int t = (int)(Math.pow(20,22)%7);
num +=t;
num%=7;
System.out.println(num+1);
思路2
你需要知道 (a * b ) % p = a % p * b % p
java
Scanner scan = new Scanner(System.in);
int num = 1;
for(int i = 1; i <= 22; i++)
// (a * b ) % p = a % p * b % p
// 这里再加上一个%p是担心 乘积会溢出
num = (num %7 * 20 % 7) % 7;
num = (num + 5) % 7 ;
System.out.println(num+1);
scan.close();
思路3
java中有一个类是专门用来进行大整数运算的:BigInteger
java
import java.math.BigInteger;
import java.text.SimpleDateFormat;
import java.time.LocalDate;
import java.util.*;
// 1:无需package
// 2: 类名必须Main, 不可修改
public class Main {
public static void main(String[] args) {
BigInteger res = new BigInteger("5");
BigInteger product = new BigInteger("20");
BigInteger twenty_two = new BigInteger("22");
BigInteger seven = new BigInteger("7");
//等同于 (20 ^22^ % 7 + 5)%7
int i = product.modPow(twenty_two, seven)
.add(res).mod(seven).intValue();
System.out.println(i+1);
}
}