LIGHT OJ 1225 - Palindromic Numbers (II)【回文数】
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1225 - Palindromic Numbers (II)
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<cstdio>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;}
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