PAT 甲级 1136 A Delayed Palindrome

1136 A Delayed Palindrome （20 point(s)）

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:

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

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.

经验总结：

emmmm 题意是，给你一个数，让你判断它是否是回文数，如果不是，进行10次转换，看能否转为回文数，转换方法是，原回文数加上逆置的原回文数，判断相加的和是否是回文数。注意数的长度可能达到1000，利用大整数加法可以轻松解决~

AC代码

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
const int maxn=1010;
int n,m,k,num;
char s[maxn];
struct bign
{
	int len,d[maxn];
	bign()
	{
		len=0;
		memset(d,0,sizeof(d));
	}
};
bool isparlindrome(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 change(char str[])
{
	bign a;
	a.len=strlen(str);
	for(int i=0;i<a.len;++i)
		a.d[i]=str[a.len-1-i]-'0';
	return a;
}
void print(bign a)
{
	for(int i=a.len-1;i>=0;--i)
		printf("%d",a.d[i]);
}
bign invert(bign a)
{
	bign b;
	b.len=a.len;
	for(int i=0;i<a.len;++i)
		b.d[i]=a.d[a.len-1-i];
	while(b.d[b.len-1]==0&&b.len>1)
		--b.len;
	return b;
}
bign add(bign a,bign b)
{
	bign c;
	c.len=max(a.len,b.len);
	int carry=0;
	for(int i=0;i<c.len;++i)
	{
		int temp=a.d[i]+b.d[i]+carry;
		c.d[i]=temp%10;
		carry=temp/10;
	}
	if(carry!=0)
		c.d[c.len++]=carry;
	return c;
}
int main()
{
	scanf("%s",s);
	bool flag=false;
	bign c=change(s);
	for(int i=0;i<=10;++i)
	{
		if(isparlindrome(c))
		{
			flag=true;
			break;
		}
		if(i==10)
			break;
		bign a=c;
		bign b=invert(c);
		print(a);printf(" + ");
		print(b);printf(" = ");
		c=add(a,b);
		print(c);printf("\n");
	}
	if(flag)
	{
		print(c);
		printf(" is a palindromic number.\n");
	}
	else
		printf("Not found in 10 iterations.\n");
	return 0;
}