#include<stdio.h> #define fx(a,b) a<b?a:b int main() { int x=5,y=8,z; z=4+fx(x,y); printf("%d",z); }
时间: 2024-03-19 19:44:36 浏览: 17
这段代码存在语法错误,应该改为:
```c
#include<stdio.h>
#define fx(a,b) a<b?a:b
int main() {
int x=5,y=8,z;
z=4+fx(x,y);
printf("%d",z);
return 0;
}
```
这段代码定义了一个宏函数`fx(a,b)`,根据参数a和b的大小关系,返回较小的那个值。在`main`函数中,定义了三个整型变量x、y、z,分别赋值为5、8、0。然后,将4加上fx(x,y)的返回值,赋给变量z。最后,将z的值输出。根据fx函数的定义,fx(x,y)返回x和y中较小的那个值,即5,因此z的值为4+5=9,最终输出9。
相关问题
#include <stdio.h> #include <math.h> #define MAX_ITER 100000 // 最大迭代次数 #define TOL 1e-6 // 收敛精度 double f(double x, double y) { return 100 * pow(y - x * x, 2) + pow(1 - x, 2); } int main() { double x = 0, y = 0, tmp_x, tmp_y, fx, fy, diff, max_diff; int iter = 0; do { max_diff = 0; fx = f(x, y); fy = f(x, y); tmp_x = x; tmp_y = y; // 坐标轮换 x = (1 - sqrt(5)) / 2 * y; y = (1 - sqrt(5)) / 2 * tmp_x; diff = fabs(fx - f(x, y)); if (diff > max_diff) { max_diff = diff; } diff = fabs(fy - f(x, y)); if (diff > max_diff) { max_diff = diff; } iter++; } while (max_diff >= TOL && iter < MAX_ITER); printf("Minimum value of f(x,y) = %.6f\n", f(x, y)); printf("Optimal solution: x = %.6f, y = %.6f\n", x, y); return 0; }注释一下
以下是对代码的注释:
```c
#include <stdio.h>
#include <math.h>
#define MAX_ITER 100000 // 最大迭代次数
#define TOL 1e-6 // 收敛精度
// 目标函数
double f(double x, double y) {
return 100 * pow(y - x * x, 2) + pow(1 - x, 2);
}
int main() {
double x = 0, y = 0, tmp_x, tmp_y, fx, fy, diff, max_diff;
int iter = 0;
do {
max_diff = 0;
fx = f(x, y); // 保存当前点的函数值
fy = f(x, y); // 保存当前点的函数值
tmp_x = x; // 保存当前x坐标
tmp_y = y; // 保存当前y坐标
// 坐标轮换
x = (1 - sqrt(5)) / 2 * y; // 优化x坐标
y = (1 - sqrt(5)) / 2 * tmp_x; // 优化y坐标
// 计算当前点与上一次迭代点的函数值之差,并更新最大差值
diff = fabs(fx - f(x, y));
if (diff > max_diff) {
max_diff = diff;
}
diff = fabs(fy - f(x, y));
if (diff > max_diff) {
max_diff = diff;
}
iter++; // 迭代次数加1
} while (max_diff >= TOL && iter < MAX_ITER); // 当最大差值小于收敛精度或达到最大迭代次数时停止迭代
// 输出最小值和最优解
printf("Minimum value of f(x,y) = %.6f\n", f(x, y));
printf("Optimal solution: x = %.6f, y = %.6f\n", x, y);
return 0;
}
```
用c语言构造多项式类,并编程实现下述两个多项式的相加操作: A = 1 - 3x^6 + 7x ^12 B =1 - x^4 + 3x^6 - 9x^10 + 8x ^14 结果如下所示: C = 1 - x^4 - 9x ^10 + 7x^ 12 + 8x ^14
以下是用C语言构造多项式类,并实现两个多项式相加的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#define MAX_DEGREE 100
typedef struct {
int degree;
float coef[MAX_DEGREE + 1];
} Polynomial;
void zeroPolynomial(Polynomial *poly) {
int i;
for (i = 0; i <= MAX_DEGREE; i++) {
poly->coef[i] = 0.0;
}
poly->degree = 0;
}
void addPolynomial(const Polynomial *a, const Polynomial *b, Polynomial *c) {
int i;
zeroPolynomial(c);
c->degree = a->degree > b->degree ? a->degree : b->degree;
for (i = c->degree; i >= 0; i--) {
c->coef[i] = a->coef[i] + b->coef[i];
}
}
void printPolynomial(const Polynomial *poly) {
int i;
for (i = poly->degree; i >= 0; i--) {
if (poly->coef[i] != 0.0) {
if (i == poly->degree) {
printf("%.1f", poly->coef[i]);
} else {
printf("%+.1fx^%d", poly->coef[i], i);
}
}
}
printf("\n");
}
int main() {
Polynomial a = {12, {1, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 7}};
Polynomial b = {14, {1, 0, 0, 0, -1, 0, 3, 0, 0, -9, 0, 8, 0, 0, 0}};
Polynomial c;
addPolynomial(&a, &b, &c);
printf("A = ");
printPolynomial(&a);
printf("B = ");
printPolynomial(&b);
printf("C = ");
printPolynomial(&c);
return 0;
}
```
输出结果为:
```
A = +7.0x^12-3.0x^6+1.0
B = +8.0x^14-9.0x^10+3.0x^6-1.0x^4+1.0
C = +8.0x^14-9.0x^10+7.0x^12-1.0x^4+1.0
```