1136. A Delayed Palindrome (20)

深碍√TFBOYSˉ_ 2022-05-28 10:40 147阅读 0赞

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:**

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.

--------------------

题目大意：

代码：

#include<stdio.h>
    #include<string.h>
    char str[10010];
    char str1[10010];
    char str2[10010];
    char str3[10010];
    int main()
    {
        int i,j,n,m,k,t,l,v,flag;
        scanf("%s",str);
        for(i=0;i<10;i++)
        {
            l=0;
            for(j=strlen(str)-1;j>=0;j--)
            {
                str1[l++]=str[j];
            }
            str1[l]='\0';
            if(strcmp(str1,str)==0)
            {
                printf("%s is a palindromic number.",str);
                return 0;
            }
            else
            {
                v=0;
                flag=0;
                for(j=strlen(str)-1,l=strlen(str1)-1;j>=0&&l>=0;j--,l--)
                {
                    int a=str[j]-'0';
                    int b=str1[l]-'0';
                    if(flag==1)
                        a++;
                    if(a+b>=10)
                    {
                    flag=1;
                    }
                    else
                    {
                       flag=0;
                    }
                    str2[v++]=((a+b)%10+'0');
                }
                while(j>=0)
                {
                    if(flag==1)
                    {
                        flag=0;
                        str[j]=str[j]+1;
                    }
                    str2[v++]=str[j--];
                }
                while(l>=0)
                {
                    if(flag==1)
                    {
                        flag=0;
                        str1[l]=str1[l]+1;
                    }
                    str2[v++]=str1[l--];
                }
                if(flag==1)
                {
                    str2[v++]='1';
                }
                str2[v]='\0';
                l=0;
                for(j=strlen(str2)-1;j>=0;j--)
                {
                    str3[l++]=str2[j];
                }
                str3[l++]='\0';
                printf("%s + %s = %s\n",str,str1,str3);
                strcpy(str,str3);
            }
        }
        printf("Not found in 10 iterations.\n");
        return 0;
    }