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  • Hdu 1266 Reverse Number

    Reverse Number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 9927    Accepted Submission(s): 4410

    Problem Description

    Welcome 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.

    Input

    Input 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.

    Output

    For each test case, you should output its reverse number, one case per line.

    Sample Input

    3

    12

    -12

    1200

    Sample Output

    21

    -21

    2100

    #include<stdio.h>  
    #include<string.h>  
    #define MAXN 10000  
    char a[MAXN];  
    int main()  
        {  
            int t, n, i, len;  
            int k;  
            scanf("%d", &t);  
            while( t-- )  
            {  
                scanf("%s", &a);  
                len = strlen(a);  
                for(k = len - 1; k >= 0; k--)  
                {  
                    if(a[k] != '0')  
                        break;  
                }  
                if(a[0] == '-')  
                {  
                    printf("-");  
                    for(i = k; i >= 1; i--)  
                        printf("%c", a[i]);  
                    for(i = k + 1; i < len; i++)  
                        printf("0");  
                }  
                else  
                {  
                    for(i = k; i >= 0; i--)  
                        printf("%c", a[i]);  
                    for(i = k + 1; i < len; i++)  
                        printf("0");  
                }  
                printf("
    ");  
            }  
            return 0;  
    }  
    

      

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  • 原文地址:https://www.cnblogs.com/zhangliu/p/7058004.html
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