@[Zehao](/space/show?uid=47312) 最好这样做：
```cpp
时间差：
int min,min1,min2;
min1 = a*60+b;
min2 = c*60+d;
min = min2-min1;
```
by zxz_sword @ 2019-05-26 18:27:33
@[Zehao](/space/show?uid=47312)
```cpp
区分小时和分钟
int hour;
hour = minute/60;
minute = minute%60;
```
by zxz_sword @ 2019-05-26 18:28:19
@[Zehao](/space/show?uid=47312) 顺便求关注
by zxz_sword @ 2019-05-26 18:28:30
@[zxz_sword](/space/show?uid=213272) 好的哥
不过我想知道我那一堆代码的问题在哪？？？
by Zehao @ 2019-05-26 18:34:09
@[Zehao](/space/show?uid=47312)
```
cout<<"Enter the time:\n";
```
????
by hfee @ 2019-05-26 18:38:05
@[白狼与玫瑰](/space/show?uid=123936)
哈哈哈我闲的没事加这一行干什么，多谢大佬指点，原来删去那一行就好了。。。
by Zehao @ 2019-05-26 18:41:11
把第一个输出去掉就行了。
by gcl2004 @ 2019-05-26 19:35:08
看看我这个c语言应该是没问题的吧
# include <stdio.h>
int main(void)
{
int a, b, c, d, sum, val;
printf("请输入小鱼的开始时间：");
scanf("%d %d", &a, &b);
printf("请输入结束时间：");
scanf("%d %d", &c, &d);
if (a < 24 && c<24 && a < c)
{
if (b<60 && d<60 && d == b)
{
printf("游了%d小时\n", c-a);
}
else if(b < d)
{
printf("游了%d小时, %d 分钟", c-a, d-b);
}
else if (b > d )
{
val = c-a-1;
sum = (60-(b-d));
printf("游了%d个小时， %d分钟\n", val, sum);
}
}
else
printf("小鱼冤死！");
return 0;
}
by yzcaaa @ 2019-06-10 16:35:24
#include<iostream>
using namespace std;
int main()
{
int a,b,c,d;
cin>>a>>b>>c>>d;
int e=a*60+b;
int f=c*60+d;
int g=f-e;
cout<<g/60<<" "<<g%60<<endl;
return 0;
}
by 养生莫熬夜 @ 2019-08-14 09:36:26
这样做可得AC哦
by 养生莫熬夜 @ 2019-08-14 09:37:03