Time Convertion

Given a time in AM/PM format, convert it to military (24-hour) time.

Note: Midnight is 12:00:00AM on a 12-hour clock and 00:00:00 on a 24-hour clock. Noon is 12:00:00PM on a 12-hour clock and 12:00:00 on a 24-hour clock.

Input Format

A time in 12-hour clock format (i.e.: hh:mm:ssAM or hh:mm:ssPM), where 01≤hh≤12.

Output Format

Convert and print the given time in 24-hour format, where 00≤hh≤23.

Sample Input

07:05:45PM
Sample Output

19:05:45

这么简单的题目，居然一直过不了， 这是自己写的代码：

#include <math.h>#include <stdio.h>#include <string.h>#include <stdlib.h>int main(){    char* time = (char *)malloc(10240 * sizeof(char));    scanf("%s",time);    if(time[strlen(time) - 2] == 'A')        for(int i = 0; i < strlen(time) - 2; i++)            printf("%c", time[i]);    else{        char hour[3];        if(time[0] != '0')            for(int i = 0; i < 2; i++)                hour[i] = time[i];        else hour[0] = time[1];        hour[2] = '\0';        int num = atoi(hour);        num += 12;        if(num == 24){            time[0] = '0';            time[1] = '0';        }        else{            sprintf(hour, "%d", num);            for(int i = 0; i < 2; i++)                time[i] = hour[i];        }        for(int i = 0; i < strlen(time) - 2; i++)            printf("%c", time[i]);    }    return 0;}

这是正确代码：(被自己蠢哭。。）

#include<iostream>#include<cstdio>using namespace std;int main(){       string s;    cin>>s;    int n=s.length();    int hh,mm,ss;    hh=(s[0]-'0')*10+(s[1]-'0');   //remember this method    mm=(s[3]-'0')*10+(s[4]-'0');    ss=(s[6]-'0')*10+s[7]-'0';    if(hh<12&&s[8]=='P')       hh+=12;    if(hh==12 && s[8]=='A')  //mess up this difference        hh=0;    printf("%02d:%02d:%02d\n",hh,mm,ss);    return 0;}