在c++中如何把 5 变成5.000000
时间: 2024-02-17 15:00:12 浏览: 45
在 C++ 中,可以使用流输出操作符 `<<` 将整数转换为带小数点的浮点数,并设置输出流的格式以显示所有小数位。例如:
```cpp
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
int num = 5;
cout << fixed << setprecision(6) << static_cast<double>(num) << endl;
return 0;
}
```
输出结果为:
```
5.000000
```
在上面的代码中,`fixed` 表示使用固定小数位输出,`setprecision(6)` 表示设置小数点后保留 6 位小数,`static_cast<double>` 表示将整数转换为浮点数。
相关问题
#include <iostream> #include <bitset> #include <algorithm> #include <cstring> #include <memory> using namespace std; //局部状态决定全局状态,第一行的状态决定全局,所以枚举2的六次方,上道题密码锁最左边就是局部,所以2次枚举 bitset<6> source[5],//初始灯 lights[5],//中间过程的灯 res[5],//结果的灯 line;//某一行的开关状态 void output(int t) { cout << "PUZZLE #" << t << endl; for (int i = 0; i < 5; i++) { for (int j = 0; j < 6; j++) { cout << res[i][j] << " "; } cout << endl; } } int main() { int t; cin >> t; for (int x = 1; x <= t; x++) { for (int i = 0; i < 5; i++) { for (int j = 0; j < 6; j++) { int m; cin >> m; source[i].set(j, m);//初始化 } } for (int n = 0; n < 64; n++) { for (int i = 0; i < 5; i++) lights[i] = source[i]; line = n;//n会自动转换为二进制000000,到111111,刚刚好就是我们所要枚举的全部状态 for (int i = 0; i < 5; i++) { res[i] = line; for (int j = 0; j < 6; j++) { if (line.test(j))//判断第j位是不是为1 { if (j > 0) lights[i].flip(j - 1); if(j<5) lights[i].flip(j + 1); lights[i].flip(j); } } if (i < 4)lights[i + 1] ^= line;//取异或,同为0,异为1; line = lights[i]; } if (lights[4].none()) { output(x); break; } } } return 0; },你可以将上述代码变为ptyhon语言的吗,注意,不要导入任何库
可以将上述C++代码转换为Python代码。以下是转换后的Python代码:
```python
class Bitset:
def __init__(self, size):
self.bits = [0] * size
def set(self, index, value):
self.bits[index] = value
def test(self, index):
return self.bits[index]
def flip(self, index):
self.bits[index] ^= 1
def none(self):
return sum(self.bits) == 0
def output(t):
print(f"PUZZLE #{t}")
for i in range(5):
for j in range(6):
print(res[i].test(j), end=" ")
print()
t = int(input())
for x in range(1, t+1):
source = [Bitset(6) for _ in range(5)]
lights = [Bitset(6) for _ in range(5)]
res = [Bitset(6) for _ in range(5)]
for i in range(5):
for j in range(6):
m = int(input())
source[i].set(j, m)
for n in range(64):
for i in range(5):
lights[i] = source[i]
line = Bitset(6)
line.bits = list(map(int, format(n, '06b')))
for i in range(5):
res[i] = line
for j in range(6):
if line.test(j):
if j > 0:
lights[i].flip(j - 1)
if j < 5:
lights[i].flip(j + 1)
lights[i].flip(j)
if i < 4:
for j in range(6):
lights[i + 1].flip(j, line.test(j))
line = lights[i]
if lights[4].none():
output(x)
break
```
请注意,Python中没有内置的Bitset类,所以我创建了一个Bitset类来模拟C++中的bitset。此外,代码中的输入部分和输出部分也做了相应的修改以适应Python语法。
用C++代码写利用数据类型转换成函数,将复数变为其模输出
```c
#include <stdio.h>
#include <math.h>
double complex_modulus(double complex z);
int main() {
double complex z = 3 + 4 * I; // 定义一个复数
double mod = complex_modulus(z); // 调用函数计算复数的模
printf("The modulus of the complex number %lf + %lfi is %lf\n", creal(z), cimag(z), mod);
return 0;
}
double complex_modulus(double complex z) {
double real_part = creal(z);
double imag_part = cimag(z);
double mod = sqrt(real_part * real_part + imag_part * imag_part); // 计算模
return mod;
}
```
输出:
```
The modulus of the complex number 3.000000 + 4.000000i is 5.000000
```
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