链接：https://ac.nowcoder.com/acm/contest/5667/D
来源：牛客网

题目描述:

Given two moments on the same day in the form of HH:MM:SS, print the number of seconds between the two moments.

输入描述：

Input two lines each contains a string in the form of HH:MM:SS (00≤HH≤23,00≤MM,SS≤59), denoting a given moment.

输出描述：

Only one line containing one integer, denoting the answer.

solution：

将时间全部化成秒，然后两个时间相减，取绝对值

#include <iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<map>
#include<queue>
#include<vector>
#include<algorithm>
#define INF 0x3f3f3f3f

using namespace std;
typedef long long ll;
typedef pair<int,int>P;
int h1,h2,m1,m2,s1,s2;

int main()
{
    scanf("%d:%d:%d",&h1,&m1,&s1);
    scanf("%d:%d:%d",&h2,&m2,&s2);
    int h,m,s;
        h=h1-h2;
        m=m1-m2;
        s=s1-s2;

    int ans=h*3600+m*60+s;
    cout<<abs(ans)<<endl;
    return 0;
}