1024 Palindromic Number (25 分)大整数相加+会问+reverse倒置

1024 Palindromic Number (25 分)

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

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. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (≤1010) is the initial numer and K (≤100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3

注意大整数乘法是bign和int以及while

大整数加法是bign和bign以及if

#include <cstdio>
#include <algorithm>#include <cstring>
using namespace std;struct bign{int d[100];int len;bign(){memset(d,0,sizeof(d));len=0;}
};//下面将整数转换成bign,逆序存放
bign change(char str[]){bign a;a.len=strlen(str);for(int i=0;i<a.len;i++){a.d[i]=str[a.len-i-1]-'0';}return a;
}//下面为高精度乘法bign add(bign a,bign b){bign c;int carry=0;//carry为进位for(int i=0;i<a.len||i<b.len;i++){int temp=a.d[i]+b.d[i]+carry;c.d[c.len++]=temp%10;carry=temp/10;}if(carry!=0){//注意这里时ifc.d[c.len++]=carry;}return c;
}bool judge(bign a){for(int i=0;i<a.len/2;i++){if(a.d[i]!=a.d[a.len-1-i])return false;}return true;
}void print(bign a){for(int i=a.len-1;i>=0;i--){printf("%d",a.d[i]);}printf("\n");
}int main(){char str[100];int t;//操作次数的上限int k=0;scanf("%s %d",str,&t);bign a=change(str);for(int i=0;i<t;i++){if(judge(a)==true){print(a);printf("%d\n",k);return 0;}//注意输入的str就是回文这个点bign b=a;reverse(b.d,b.d+b.len);a=add(a,b);k++;}print(a);printf("%d\n",k);return 0;}