9 条题解
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4Atterger(liyuwen) LV 5 @ 2023-5-11 21:43:45
这题直接用
printf
里面的%.2f
就行,代码如下~#include <bits/stdc++.h> using namespace std; int main() { double x, y, temp = 0; cin >> x >> y; temp = x * y; printf ("%.2f",temp); return 0; }
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22023-8-15 16:52:25@
/********************************* 备注: *********************************/ #include <iostream> #include <cstdio> #include <iomanip> #include <cmath> #include <math.h> #include <algorithm> #include <cstring> #include <string> #include <stack> #include <queue> #include <algorithm> #define LL long long using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5 + 10; double n , m; int main() { cin >> n >> m; double zzz = n * m; cout << fixed << setprecision(2) << zzz << endl; return 0; }
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12023-1-23 15:10:12@
#include <iostream> using namespace std; int main(){ double a,b; scanf("%lf%lf",&a,&b); printf("%.2lf",a*b); return 0; }
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12022-8-24 22:24:45@
#include <bits/stdc++.h> using namespace std; int main(void) { float x, y; cin >> x >> y; printf("%.2f", x * y); return 0; }
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12022-1-25 10:15:57@
#include <stdio.h> #include <iostream> using namespace std; int main() { double x,y; cin >> x >> y; printf("%.2lf",x*y); }
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-12022-9-23 21:19:19@
#include<stdio.h> #include<iostream> using namespace std; int main() { double x,y; cin >> x >> y; printf("%.2lf,x*y);
return 0;
}
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-12022-7-2 10:01:48@
#include <stdio.h> #include <iostream> using namespace std; int main() { double x,y; cin >> x >> y; printf("%.2lf",x * y); }
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-12022-7-1 18:13:48@
#include<stdio.h> #include<iostream> #include<iomanip> using namespace std; int main() { double x,y; cin>>x>>y; cout<<fixed<<setprecision(2)<<x*y<<endl; }
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-22022-1-21 23:02:34@
#include <iostream> #include <iomanip> using namespace std; int main(){ double x,y; cin>>x>>y; cout<<fixed<<setprecision(2)<<x*y<<endl; return 0; }
- 1
信息
- ID
- 828
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 3
- 标签
- 递交数
- 790
- 已通过
- 412
- 上传者