Getit(int m) { Int I,k; K = sqrt( m ); For( I = 2;i<=k;i++){ If( m%i==0) break; If( i>=k+1 ) Printf(“a”) } Else Printf(“b”) } 对以上c语言代码,请写出满足 100% DC (判定覆盖)所需的逻辑条件。
时间: 2023-12-07 08:02:42 浏览: 50
这段代码有语法错误,else语句的位置不正确,应该放在if语句的后面,所以无法进行100%的判定覆盖。以下是修改后的代码及满足100% DC的逻辑条件:
```
Getit(int m) {
int i, k;
k = sqrt(m);
for (i = 2; i <= k; i++) {
if (m % i == 0) {
break;
}
}
if (i >= k + 1) {
printf("a");
} else {
printf("b");
}
}
```
满足100% DC的逻辑条件:
- m为正整数
- k为m的平方根取整后的结果
- i初始值为2
- 循环条件为i<=k
- 如果m能被i整除,则执行break跳出循环
- 如果i>=k+1,则输出字符a
- 否则,输出字符b
相关问题
用帅逸法查找2到100之内的素数,(请在空中填写表达式或语句等)include (ath h> Bincluda (andio.h)roid ged(int *a ) int i,j For(i-2; (100:5+)) a[i7-_ . /9用 下标对应数组元素值 来初始化数组*/for(i=2: i<sqrt(100);it+) for(j=___ :j<=100:j++) if(a[i]!= 0 8 a([j]_____) if(a[j]%a[i]== 0) a[j_____ ; void main( ( int i, a[101]={ .) /* 数组初始化为0 */ gcd( a ); /*函数调用, 数组名作为实参*/ for(i=2; i<=100;i++) if(a[i]!=0) printf("%4d",a[i]);
#include <stdio.h>
#include <math.h>
void getPrimes(int *a) {
int i, j;
// 用下标对应数组元素值来初始化数组
for (i = 2; i <= 100; i++) {
a[i] = i;
}
// 利用筛选法求素数
for (i = 2; i <= sqrt(100); i++) {
for (j = i + 1; j <= 100; j++) {
if (a[i] != 0 && a[j] % a[i] == 0) {
a[j] = 0;
}
}
}
}
int main() {
int i, a[101] = {0};
getPrimes(a);
for (i = 2; i <= 100; i++) {
if (a[i] != 0) {
printf("%4d", a[i]);
}
}
return 0;
}
write a C example use Newton-Raphson to get Fractrals for f(z)=z^3-1=0 and save the result into "Problem1.txt"
Sure, here's a C example that uses Newton-Raphson method to get fractals for f(z) = z^3 - 1 = 0:
```c
#include <stdio.h>
#include <math.h>
#define MAX_ITER 1000
#define EPSILON 1e-6
typedef struct {
double x;
double y;
} Complex;
void newton_raphson(Complex z, Complex *z_new) {
Complex numerator = {pow(z.x, 3) + 3 * z.x * pow(z.y, 2) - 1, 3 * pow(z.x, 2) * z.y - pow(z.y, 3)};
Complex denominator = {3 * pow(z.x, 2) + 3 * pow(z.y, 2), 6 * z.x * z.y};
z_new->x = z.x - numerator.x / denominator.x;
z_new->y = z.y - numerator.y / denominator.y;
}
int main() {
FILE *fp;
fp = fopen("Problem1.txt", "w");
if (fp == NULL) {
printf("Error opening file\n");
return 1;
}
for (double x = -2; x <= 2; x += 0.01) {
for (double y = -2; y <= 2; y += 0.01) {
Complex z = {x, y};
Complex z_new = {0, 0};
int iter = 0;
while (iter < MAX_ITER) {
newton_raphson(z, &z_new);
double distance = sqrt(pow(z_new.x - z.x, 2) + pow(z_new.y - z.y, 2));
if (distance < EPSILON) {
fprintf(fp, "%.2lf %.2lf\n", x, y);
break;
}
z = z_new;
iter++;
}
}
}
fclose(fp);
return 0;
}
```
This code performs a double loop over the region from -2 to 2 in both the x and y dimensions with a step size of 0.01. For each point (x, y), the code applies the Newton-Raphson method to find the root of the equation f(z) = z^3 - 1 starting from the initial point z = (x, y). The method is iterated until either the maximum number of iterations is reached (MAX_ITER) or the distance between successive iterations is below a small tolerance (EPSILON).
If the root is found, the point (x, y) is considered to be part of the fractal and is written to the file "Problem1.txt" in the format "x y".
Note that this code makes use of complex arithmetic and is fairly computationally intensive, so it may take a while to run depending on your machine.