刚考完c语言,赶紧拿到电脑上敲一下,看看自己写的对不对。
打印半径位1-10的圆的面积,若面积超过100,则不打印
int main() {
int i = 0;
double s = 0;
for (i = 1; i <= 10; i++) {
s = 3.14 * i * i;
if (s > 100)
continue;
printf("%.2f ", s);
}
}
将以下二维数组a行列互换,存放到另一个二维数组b
[1 2 3] [1 4]
a [ ] b[2 5]
[4 5 6] [3 6]
int main() {
int a[2][3] = { {1,2,3},{4,5,6} };
int b[3][2] = {0};
int i = 0;
int j = 0;
for (i = 0; i < 2; i++) {
for (j = 0; j < 3; j++) {
b[j][i] = a[i][j];
}
}
for (i = 0; i < 3; i++) {
for (j = 0; j < 2; j++)
printf("%d ", b[i][j]);
printf("\n");
}
return 0;
}
请输入一个3*3的矩阵。输出主对角线与反对角线元素的和。
分三个函数输入,计算,输出。
void input(int arr[3][3]) {
int i = 0;
int j = 0;
printf("请输入数组arr[3][3]:\n");
for (i = 0; i < 3; i++)
for (j = 0; j < 3; j++)
scanf("%d", &arr[i][j]);
}
int Sum(int arr[3][3]) {
int sum = 0;
sum = arr[0][0] + arr[1][1] + arr[2][2] + arr[0][2] + arr[2][0];
return sum;
}
void output(int sum) {
printf("主对角线与反对角线之和为:%d\n", sum);
}
int main() {
int arr[3][3] = { 0 };
int sum = 0;
input(arr);
sum = Sum(arr);
output(sum);
return 0;
}
有两个字符串:字符串a内容为“I am a teacher.” 字符串b的内容为“You are a student.”
要求:将字符串b的内容连接到字符串a的后面,即字符串a的内容为:“I am a teacher.You are a student.”(指针实现)
int main() {
char str1[50] = "I am a teacher.";
char str2[50] = "You are a student.";
char* p1 = str1;
char* p2 = str2;
int len = 0;
len = strlen(str1);
p1 += len;
while (*p2 != '\0') {
*p1++ = *p2++;
}
*p1++ = '\0';
puts(str1);
}
思考:
考试的时候我是这么写的,唉,应该是数组越界了,拿出来挨打把!
int main() {
char str1 = "I am a teacher.";
char str2 = "You are a student.";
char* p1 = str1;
char* p2 = str2;
int len = 0;
len = strlen(str1);
p1 += len;
while (*p2 != '\0') {
*p1++ = *p2++;
}
*p1++ = '\0';
puts(str1);
}