题解

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main(){
	int n, m, k;
    int ct1, ct2;
    scanf("%d:%d:%d", &n, &m, &k);//scanf运用
	ct1 = n * 3600 + m * 60 + k;
    scanf("%d:%d:%d", &n, &m, &k);
    ct2 = n * 3600 + m * 60 + k;
    cin >> n;//循环利用
    cout << (ct2 - ct1) * n;

	return 0;
}

时间复杂度：$O(1)$