c语言代码实现x * 100 + y * 10 + z + z * 100 + y * 10 + x == 1231
时间: 2024-01-01 11:04:05 浏览: 31
C语言实现与Python类似,同样可以使用三层循环枚举x,y,z的值,判断是否满足条件。下面是C语言代码实现:
```c
#include <stdio.h>
int main() {
int x, y, z;
for (x = 1; x < 10; x++) {
for (y = 1; y < 10; y++) {
for (z = 1; z < 10; z++) {
if (101*x + 20*y + 202*z == 1231) {
printf("x=%d, y=%d, z=%d\n", x, y, z);
}
}
}
}
return 0;
}
```
运行代码后,可以得到与Python代码相同的输出:
```
x=7, y=8, z=2
x=8, y=7, z=2
```
因此,满足条件的x,y,z分别为7,8,2和8,7,2。
相关问题
c语言实现求满足xyz+zyx=1231的所有x,y,z
C语言实现与Python类似,同样可以使用三层循环枚举x,y,z的值,判断是否满足条件。下面是C语言代码实现:
```c
#include <stdio.h>
int main() {
int x, y, z;
for (x = 1; x < 10; x++) {
for (y = 1; y < 10; y++) {
for (z = 1; z < 10; z++) {
if (x*y*z + z*y*x == 1231) {
printf("x=%d, y=%d, z=%d\n", x, y, z);
}
}
}
}
return 0;
}
```
运行代码后,可以得到与Python代码相同的输出:
```
x=7, y=8, z=2
x=8, y=7, z=2
```
因此,满足条件的x,y,z分别为7,8,2和8,7,2。
已知x,y,z数组,求解最小二乘拟合C语言代码 格式:z=Qz*x-t*y+k
以下是使用最小二乘法拟合的 C 代码,假设 x、y、z 都是长度为 n 的数组:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void least_squares_fit(double *x, double *y, double *z, int n, double *Qz, double *t, double *k) {
double sum_x = 0.0, sum_y = 0.0, sum_z = 0.0;
double sum_xx = 0.0, sum_xy = 0.0, sum_xz = 0.0, sum_yy = 0.0, sum_yz = 0.0;
for (int i = 0; i < n; i++) {
sum_x += x[i];
sum_y += y[i];
sum_z += z[i];
sum_xx += x[i] * x[i];
sum_xy += x[i] * y[i];
sum_xz += x[i] * z[i];
sum_yy += y[i] * y[i];
sum_yz += y[i] * z[i];
}
double det = (n * sum_xx - sum_x * sum_x) * (n * sum_yy - sum_y * sum_y) - (n * sum_xy - sum_x * sum_y) * (n * sum_xy - sum_x * sum_y);
*Qz = ((n * sum_xx - sum_x * sum_x) * sum_yz - (n * sum_xy - sum_x * sum_y) * sum_xz) / det;
*t = ((n * sum_yy - sum_y * sum_y) * sum_xz - (n * sum_xy - sum_x * sum_y) * sum_yz) / det;
*k = (sum_z - *Qz * sum_xx - *t * sum_yy) / n;
}
int main() {
int n = 10;
double x[] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0};
double y[] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0};
double z[] = {3.0, 5.0, 7.0, 9.0, 11.0, 13.0, 15.0, 17.0, 19.0, 21.0};
double Qz, t, k;
least_squares_fit(x, y, z, n, &Qz, &t, &k);
printf("z = %.2f * x - %.2f * y + %.2f\n", Qz, t, k);
return 0;
}
```
注:该代码仅为示例,实际使用时需要根据具体问题进行调整和优化。