10 条题解
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0钟皓宇 (zhonghaoyu) LV 7 @ 2023-8-4 15:07:49
/************************************ 备注: ************************************/ #include <iostream> #include <cstdio> #include <iomanip> #include <cmath> #include <algorithm> #include <cstring> #include <string> #include <stack> #include <queue> #include <bits/stdc++.h> #define LL long long using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5 + 10; double n; int main() { cin >> n; cout << fixed << setprecision(3) << n << endl; return 0; }
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02023-6-1 21:33:38@
#include <iostream> #include <bits/stdc++.h> using namespace std; const int N=1e7+10; const int INF=0x3f3f3f3f; int main() { float a; cin>>a; cout<<fixed<<setprecision(3)<<a; }
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02023-5-12 17:42:02@
#include <iostream> using namespace std; int main() { double n; cin >> n; printf ("%.3f",n); return 0; }
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02022-10-30 16:21:55@
#include <bits/stdc++.h> using namespace std; int main(void) { float a; cin >> a; cout << fixed << setprecision(3) << a; return 0; }
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-12022-10-17 17:21:26@
这道题要多练😕
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <string> #include <iomanip> #includejf #includedkjfsj #include<;;; #include #include using namespe std; #define long long LL const int N = 1e6+6; const int INF = 0jbjb设 int maisasa int n; cin >> n; cout << n / 3600 << endl; cout << (n - 3600 * (n / 3600)) / 60 << endl; cout << n % 60; return 0; ewsafasad dsdssfs }
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-22023-1-23 15:19:09@
#include <iostream> using namespace std; int main(){ double a; scanf("%lf",&a); printf("%.3f",a); return 0; }
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-32022-7-23 21:18:53@
#include<bits/stdc++.h> #include <queue> #include <math.h> #include <stack> #include <stdio.h> #include <iostream> #include <vector> #include <iomanip> #include <string.h> #include <algorithm> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; int main() { float n; scanf("%f",&n); printf("%.3f",n); return 0; }
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-52022-7-1 19:05:08@
#include<stdio.h> #include<iostream> #include<iomanip> using namespace std; int main() { double a; cin>>a; cout<<fixed<<setprecision(3)<<a<<endl; }
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-52022-4-14 21:33:00@
#include<iostream> #include<iomanip> using namespace std; int main(){ double a; cin >> a; cout << fixed << setprecision(3) << 1.0* a << endl; }
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-92021-10-26 12:54:22@
n = eval(input()) print("%.3f" % n)
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信息
- ID
- 844
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 2
- 标签
- 递交数
- 707
- 已通过
- 411
- 上传者