zzh0118 @ 2024-03-04 20:28:40
#include<bits/stdc++.h>
using namespace std;
int cnt=480,s,u;
int main(){
cin>>s>>u;
cnt-=10;
cnt-=s/u;
if(cnt%60==0){
cout<<"0"<<cnt/60<<":"<<cnt%60;
}
else{
cout<<"0"<<cnt/60<<":"<<cnt%60-1;
}
return 0;
}
求调
by 2345A @ 2024-03-05 10:20:13
@zzh0118
#include<bits/stdc++.h>
using namespace std;
int main()
{
int s, v, x = 0, shijian = 0, xiaoshi = 0, fengzhong = 0;
cin >> s >> v;
x = ceil(1.0 * s / v) + 10;//要乘1.0
shijian = 60 * 32 - x;
xiaoshi = (shijian / 60) % 24;
fengzhong = shijian % 60;
if (xiaoshi < 10&&fengzhong>=10)
{
cout << "0" << xiaoshi << ":" << fengzhong;
}
if(xiaoshi<10&&fengzhong<10)
{
cout<<"0"<<xiaoshi<<":0"<<fengzhong;
}
if(xiaoshi>=10&&fengzhong>=10)
{
cout<<xiaoshi<<":"<<fengzhong;
}
if(xiaoshi>=10&&fengzhong<10)
{
cout<<xiaoshi<<":0"<<fengzhong;
}
return 0;
}
by 2345A @ 2024-03-05 10:21:42
@zzh0118 你的判断条件太多了
by 2345A @ 2024-03-05 10:35:23
@zzh0118 再用你的可能的思路写一个:
#include<bits/stdc++.h>
using namespace std;
int main(){
int t,n,x,y;
double s,v;//小数,小数!!!
cin>>s>>v;
t=ceil(s/v)+10;
n=8*60+24*60-t;
if(n>=24*60) n-=24*60;
x=n/60;
y=n%60;
if(x>=10){//你一般用的优良马蜂
if(y>=10) cout<<x<<":"<<y;
else cout<<x<<":"<<"0"<<y;
}
else {
if(y>=10) cout<<"0"<<x<<":"<<y;
else cout<<"0"<<x<<":"<<"0"<<y;
}
return 0;
}
by 2345A @ 2024-03-05 10:48:49
@zzh0118 最后,你前面的代码没有问题,后面的代码判断的太多了,所以我修改了你的代码主要是你的马蜂我真看不懂
#include<bits/stdc++.h>
using namespace std;
int cnt=8*60+24*60,num,fen,shi;
int main(){
double s,u;
cin>>s>>u;
num=ceil(s/u)+10;//ceil的意思是向上取整,因为C++默认向下取整;
cnt=8*60+24*60-num;
if(cnt>=24*60){
cnt-=24*60;
}
shi=cnt/60;
fen=cnt%60;
if(shi>=10)
{
if(fen>=10)
cout<<shi<<":"<<fen;
else
cout<<shi<<":0"<<fen;
}
else
{
if(fen>=10)
{
cout<<"0"<<shi<<":"<<fen;
}
else
cout<<"0"<<shi<<":0"<<fen;
}
return 0;
}