大整数在PAT中考的较少，但是也可能会考，在考PAT前，我是准备遇到了就直接Java的，毕竟记C++的写法心累 ＞﹏＜ 。话说都C++11/14/17/20了，给个高精度库行不行。这次来举一个常见的回文相关的大整数题，用的C++，实现的是相等数位的正整数相加。最后在介绍下Java的大整数类。
1136 A Delayed Palindrome (20 分)
Consider a positive integer N written in standard notation with k+1 digits a
Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )
Given any positive integer, you are supposed to find its paired palindromic number.
Input Specification:
Each input file contains one test case which gives a positive integer no more than 1000 digits.
Output Specification:
For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:
A + B = C
where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number – in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.
Sample Input 1:
97152
Sample Output 1:
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
Sample Input 2:
196
Sample Output 2:
196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.
题目意思很简单，看示例就明白了。
容我先给出代码
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
bool isPalin(string str){
string str2 = str;
reverse(str.begin(),str.end());
return str == str2;
}
string plusStr(string s1,string s2){
int c = 0;
string res;
for(int i=s1.size()-1;i>=0;i--){
int ans = c+s1[i]+s2[i]-'0';
c = 0;
if(ans > '9'){
c = 1;
ans -= 10;
}
res += ans;
}
if(c == 1)
res += "1";
reverse(res.begin(),res.end());
return res;
}
int main() {
string str;
cin >> str;
if(isPalin(str)){
cout << str << " is a palindromic number.";
return 0;
}
string res = str;
int time = 10;
while(!isPalin(res) && time--){
string temp = res;
reverse(temp.begin(),temp.end());
cout << res << " + " << temp << " = ";
res = plusStr(res,temp);
cout << res << endl;
}
if(time == -1 && !isPalin(res))
cout << "Not found in 10 iterations.";
else
cout << res << " is a palindromic number.";
return 0;
}
一、函数plusStr的实现
数位相等的数相加，只需从低位到高位相加然后再倒置（记得要处理最后一个进位c，如果有进位要在末尾+’1’）就可以了。
JAVA版
貌似其他没有什么说的，再附上Java大整数解法吧。用vscode写的，果然加上自动补全、错误提示的话，没有code::block顺畅，不过没有这些又不会写( ╯□╰ )
String是final类不能被继承且为字符串常量，而StringBuilder和StringBuffer均为字符串变量。所以要改变字符串同时操作较复杂时一般用StringBuilder，而且它的方法也多是这方面的，就比如reverse。- 背下输入吧，
Scanner input = new Scanner(System.in);，然后input.可以看到可以读取很多不同类型的信息，还有很多中输入，这种感觉挺方便的。 - 大名鼎鼎的
BigInteger，背下吧，记住Java没有运算符重载，所以所有常用的操作都是以方法的形式。如.add() System.out.print不带回车，System.out.println带回车，- 字符串比较用
.equals()，Java里面可以直接字符串相加，甚至可以字符串加上其他类型（相当于那个类型调用了toString()），
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String str = input.next();
int time = 10;
for (int i = 0; i < time; i++) {
StringBuilder temp = new StringBuilder(str);
String str2 = temp.reverse().toString();
if(str.equals(str2)){
System.out.print(str+" is a palindromic number.");
return;
}
BigInteger b1 = new BigInteger(str);
System.out.print(b1);
b1 = b1.add(new BigInteger(str2));
str = b1.toString();
System.out.println(" + "+str2+" = "+str);
}
System.out.print("Not found in 10 iterations.");
}
}
总结，作者水平有限，有哪里有问题都欢迎评论指出。如果对你有帮助欢迎点赞。
