LIGHT OJ 1225 - Palindromic Numbers (II)【回文数】

1225 - Palindromic Numbers (II)

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

A palindromic number or numeral palindrome is a 'symmetrical' number like 16461, that remains the same when its digits are reversed. In this problem you will be given an integer, you have to say whether the number is a palindromic number or not.

Input

Input starts with an integer T (≤ 20000), denoting the number of test cases.

Each case starts with a line containing an integer n (0 ≤ n < 10⁹).

Output

For each case, print the case number and 'Yes' if n is palindromic, otherwise print 'No'.

Sample Input	Output for Sample Input
5 1 21 16161 523125 0	Case 1: Yes Case 2: No Case 3: Yes Case 4: No Case 5: Yes

 

题意：判断所给数是否是对称的；

思路：将数的每一位数字都保存成字符，一位对比两个比较两个字符串是否相等；

失误：这种题数据那么小，最多只有9位数，就不要再考虑什么技巧判断字符串长度的奇偶性什么的，考虑越多失误越多，在比赛时浪费时间越多，应该充分利用题目的条件，有时候简单粗暴一点会很好。

AC代码：

#include

int main()
{
	int T,i,j,Kase=0,N;
	char str[100];
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&N);
		i=0;
		while(N)
		{
			str[++i]=N%10;
			N/=10;
		}
		for(j=1;j<=i;++j)
		{
			if(str[j]!=str[i-j+1])
			break;
		}
		printf("Case %d: ",++Kase);
		if(j==i+1) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}