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.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn=1010;
struct bign{
int d[maxn];
int len;
bign()
{
memset(d,0,sizeof(d));
len=0;
}
};
bign change(char str[])
{
int len=strlen(str);
bign a;
a.len=len;
for(int i=0;i<len;i++)
{
a.d[i]=(str[len-i-1]-'0');
}
return a;
}
bool isP(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;
}
bign add(bign a)
{
bign c;
int up=0;
c.len=a.len;
for(int i=0;i<a.len;i++)
{
up+=a.d[i]+a.d[a.len-1-i];
c.d[i]=up%10;
up/=10;
}
if(up)c.d[c.len++]=up;
return c;
}
bign rever(bign a)
{
bign b;
b.len=a.len;
for(int i=0;i<a.len;i++)
{
b.d[b.len-1-i]=a.d[i];
}
return b;
}
void printbign(bign a)
{
for(int i=a.len-1;i>=0;i--)
printf("%d",a.d[i]);
}
int main()
{
char str[maxn];
gets(str);
bign a=change(str);
int k=0;
while(1)
{
if(isP(a)==true)
{
printbign(a);
printf(" is a palindromic number.\n");
break;
}else if(k==10)
{
printf("Not found in 10 iterations.\n");
break;
}
printbign(a);
printf(" + ");
printbign(rever(a));
printf(" = ");
a=add(a);
printbign(a);
printf("\n");
k++;
}
return 0;
}