1136. A Delayed Palindrome (20)
Consider a positive integer N written in standard notation with k+1 digits ai as ak...a1a0 with 0 <= ai < 10 for all i and ak> 0. Then N is palindromic if and only if ai = ak-i for all i. Zero is written 0 and is also palindromic by definition.
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 1:97152Sample Output 1:
97152 + 25179 = 122331 122331 + 133221 = 255552 255552 is a palindromic number.Sample Input 2:
196Sample 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.
题意：给一个不超过一千位的整数，把他与他的镜像数字相加，直到和为一个对称的数字为止。加十次如果还不是对称的数字也退出
思路：一千位的话直接用字符串读入记为a，每次循环开始检查a是否左右对称，然后fc函数里先求出他的镜像数字b，因为a和b是左右相反的，所以直接从左往右加到ans里面就可以，jin用来标记是否进位，最后需要再检查一下最前面是否多了一位，也就是加完之后jin是否为1。此时ans是反相的，倒过来复制给a就好，同时刚好可以单个字符输出
#include<cstdio>
#include<cstring>using namespace std;char a[3000],b[3000],ans[3000];bool isans(){int n=strlen(a);int i;bool f=true;for(i=0;i*i<=n;i++){if(a[i]!=a[n-1-i]){f=false;break;}}return f;
}void fc(){int n=strlen(a);int i,j;for(i=0;i<n;i++){b[i]=a[n-1-i];}b[i]='\0';printf("%s + %s = ",a,b);int jin=0,t;for(j=0;j<n;j++){t=a[j]-'0'+b[j]-'0'+jin;if(t>=10){jin=1;t=t%10;}else{jin=0;}ans[j]=t+'0';}if(jin==1){ans[j]='1';j++;}ans[j]='\0';n=strlen(ans);for(i=0;i<n;i++){printf("%c",ans[n-1-i]);a[i]=ans[n-1-i];}a[i]='\0';printf("\n");
}int main(){int i=0;scanf("%s",a);for(i=0;i<10;i++){if(isans()){break;}fc();}if(i<10){printf("%s is a palindromic number.",a);}else{printf("Not found in 10 iterations.");}return 0;
}