【PAT甲级 BigInteger运算】1024 Palindromic Number (25 分) Java 全部AC

题目

题解 Java

一开始用C++写的，有两个大数的测试点过不去，后来改用Java的BigInteger开挂，全部通过

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;public class Main {public static void main(String[] args) throws IOException {BufferedReader br = new BufferedReader(new InputStreamReader(System.in));String str = br.readLine();String[] arr = str.split(" ");BigInteger num = new BigInteger(arr[0]);int step = Integer.parseInt(arr[1]);// 翻转for (int i = 0; i < step; i++) {BigInteger tmp = new BigInteger("" + num);BigInteger rev = new BigInteger("" + 0);while (!(num.compareTo(BigInteger.ZERO) == 0)) {// 判断是否为0rev = rev.multiply(new BigInteger("" + 10));rev = rev.add(num.mod(new BigInteger("" + 10)));num = num.divide(new BigInteger("" + 10));}// 判断是否回文:翻转后相同//System.out.println(tmp + "---" + rev);if (tmp.compareTo(rev) == 0) {System.out.println(rev + "\n" + i);return;} else {// 相加num = tmp.add(rev);}}System.out.println(num + "\n" + step);}
}

题解 C++

有两个测试点过不去

#include<iostream>
using namespace std;
int main() {long long int num, step;cin >> num >> step;//翻转for (int i = 0; i < step; i++) {long long int tmp = num;long long int rev = 0;while (num != 0) {rev *= 10;rev += num % 10;num /= 10;}//判断是否回文:翻转后相同if (tmp == rev) {cout << rev << "\n" << i;return 0;}else {//相加num = tmp + rev;}}cout << num <<"\n"<< step;system("pause");
}

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【PAT甲级 BigInteger运算】1024 Palindromic Number (25 分) Java 全部AC

题目

题解 Java

题解 C++

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