@[yatloimly](/user/1418976) 圆的面积公式是$S=πr^2$ ~~（建议重修小学数学）~~ 。求关
by fantangsong @ 2024-08-19 21:59:49
@[yatloimly](/user/1418976) `cout<<fixed<<setprecision(4)<<y*y*x;`
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不是圆的面积是 $\pi^2r$ ？
by dongzirui0817 @ 2024-08-19 22:00:55
**圆面积公式错了。。。话说你都不测一下样例吗**
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
double x,y=3.14159;
cin >> x;
cout << fixed << setprecision(4) << x*2 << " ";
cout << fixed << setprecision(4) << 2*x*y << " ";
cout << fixed << setprecision(4) << y*x*x;
return 0;
}
```
by wubisheng @ 2024-08-19 22:01:12
@[fantangsong](/user/737532) @[dongzirui0817](/user/905593) @[wubisheng](/user/893238) 哦，对，蟹蟹
by yatloimly @ 2024-08-19 22:04:07
@[yatloimly](/user/1418976) [给你的东西（已记载）](https://www.luogu.com.cn/article/ivh8q1c4)
by are_you_sure @ 2024-08-19 22:11:00
@[are_you_sure](/user/1367333) 啊这
by yatloimly @ 2024-08-19 22:31:24
@[yatloimly](/user/1418976)
可以用printf求小数位：printf("%.[保留的小数位]lf",[要被保留的算式]);
如：printf("%.4lf ",2*a); 把2*a的值保留4位小数；
~~回归真题：~~
------------
在教你define:#define [要定义的] [本身的]
如#define PI 3.14159//以后的3,14159都用PI来代替；
------------
回归正题：你的问题是公式用错了，应该是PI pow（r,r） _pow(r,r) //r的平方_
试试这个：//```cpp
```cpp
#include <iostream>
#include <cmath>
#define PI 3.14159
using namespace std;
double a;
int main(){
cin>>a;
printf("%.4lf ",2*a);
printf("%.4lf ",2*PI*a);
printf("%.4lf",PI*a*a);
return 0;
}
by youthiscoming @ 2024-08-20 08:52:00
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by AAHFXAA @ 2024-08-30 17:59:21