根本不需要那么麻烦，用scanf输入就可以了

#include <bits/stdc++.h>
using namespace std;
int h,m,s,a,b,c;
int main()
{
    scanf("%d:%d:%d\n%d:%d:%d",&h,&m,&s,&a,&b,&c);
    printf("%d",(h*3600+m*60+s)-(a*3600+b*60+c));
    return 0;
}

#include <iostream>
using namespace std;
int ha,hb,ma,mb,seca,secb,ta,tb,pos;
string sa,sb;
int main(){
    cin>>sa>>sb;
    ha=(sa[0]-'0')*10+(sa[1]-'0');
    ma=(sa[3]-'0')*10+(sa[4]-'0');
    seca=(sa[6]-'0')*10+(sa[7]-'0');
    hb=(sb[0]-'0')*10+(sb[1]-'0');
    mb=(sb[3]-'0')*10+(sb[4]-'0');
    secb=(sb[6]-'0')*10+(sb[7]-'0');    
    ta=ha*3600+ma*60+seca;
    tb=hb*3600+mb*60+secb;
    cout<<ta-tb<<endl;
    return 0;
}

#include <iostream> #include <string> using namespace std; int main() { string str1, str2; getline(cin, str1); getline(cin, str2); int str1_h = ((str1[0] - 48) * 10 + (str1[1] - 48)) * 3600;//时->秒 int str1_m = ((str1[3] - 48) * 10 + (str1[4] - 48)) * 60;//分->秒 int str1_s = ((str1[6] - 48) * 10 + (str1[7] - 48));//秒 int str1_sum = str1_h + str1_m + str1_s; int str2_h = ((str2[0] - 48) * 10 + (str2[1] - 48)) * 3600;//时->秒 int str2_m = ((str2[3] - 48) * 10 + (str2[4] - 48)) * 60;//分->秒 int str2_s = ((str2[6] - 48) * 10 + (str2[7] - 48));//秒 int str2_sum = str2_h + str2_m + str2_s; cout << str1_sum - str2_sum; return 0; }

只要先输入各自的时、分、秒，再统一计算成秒，相减即可 #include <bits/stdc++.h> using namespace std; int a, b, c, sum1, x, y, z, sum2; char t; int main() { cin >> a >> t >> b >> t >> c; cin >> x >> t >> y >> t >> z; sum1 = a * 3600 + b * 60 + c; sum2 = x * 3600 + y * 60 + z; cout << sum1 - sum2; return 0; }

#include <bits/stdc++.h>
using namespace std;
int h,m,s,a,b,c;
int main()
{
    scanf("%d:%d:%d\n%d:%d:%d",&h,&m,&s,&a,&b,&c);
    printf("%d",(h*3600+m*60+s)-(a*3600+b*60+c));
    return 0;
}

#include<bits/stdc++.h>
using namespace std;int h,m,s,a,b,c;int main(){scanf("%d:%d:%d\n%d:%d:%d",&h,&m,&s,&a,&b,&c);printf("%d",(h*3600+m*60+s)-(a*3600+b*60+c));}