Hduoj1266【水题】

/*Reverse NumberTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5182    Accepted Submission(s): 2412Problem DescriptionWelcome to 2006'4 computer college programming contest!Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100. InputInput file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow. OutputFor each test case, you should output its reverse number, one case per line. Sample Input312-121200 Sample Output21-212100 Authorlcy SourceHDU 2006-4 Programming Contest */#include<stdio.h>#include<string.h>int main(){char s[25],ch;__int64 i, j, k, m ,n, sum;scanf("%I64d",&n);getchar();while(n--){sum = 0;gets(s);m = strlen(s);if( strchr(s, '-') ){k = 0;j = 0;for(i = 1; i < m; i++){if(s[i] != '0' && !k )k = i;if(s[m-i] != '0' && !j )j = m-i;if( k && j)break;}for(i = k; i <= (j+k)/2; i++){ch = s[i];s[i] = s[ j+k - i];s[j+k-i] = ch;}if(!k)k = 1;for(i = k; i < m; i++){sum = sum * 10 + (s[i] - 48);}if(sum == 0)printf("0\n");elseprintf("-%I64d\n", sum);}else{k = -1;j = -1;for(i = 0; i < m; i++){if(s[i] != '0' && k == -1 )k = i;if(s[m-1-i] != '0' && j == -1)j = m-1-i;if( k != -1 && j != -1)break;}for(i = k; i <= (j+k)/2 ; i++){ch = s[i];s[i] = s[ j+k - i];s[j+k-i] = ch;}if(k == -1)k = 0;for(i = k; i < m; i++){sum = sum * 10 + (s[i] - 48);}printf("%I64d\n", sum);}}return 0;}