求毕达哥拉斯三元组的c程序 inta、b、n; printf(“输入自然数n(n
在嵌套的代码块周围放上大括号求毕达哥拉斯三元组的c程序 inta、b、n; printf(“输入自然数n(n,c,pythagorean,C,Pythagorean,在嵌套的代码块周围放上大括号 int a,b,n; printf("Input Natural Number n (n<2,100,000,000) : "); scanf("%d",&n); for(a=1;a<=100;a++) for(b=1;b<=100;b++) if(a<b && a*a + b*b == n*n) { printf("(%d, %d, %d)\n",
int a,b,n;
printf("Input Natural Number n (n<2,100,000,000) : ");
scanf("%d",&n);
for(a=1;a<=100;a++)
for(b=1;b<=100;b++)
if(a<b && a*a + b*b == n*n)
{
printf("(%d, %d, %d)\n",a,b,n);
}
/*else
{
printf("impossible \n");
}
*/
return 0;
inta、b、n;
int=1;
printf(“输入自然数n(n)这里是一个可能的答案
int a, b, n;
int impossible = 1;
printf("Input Natural Number n (n<2,100,000,000) : ");
scanf("%d", &n);
for (a = 1; a <= 100; a++) {
for (b = 1; b <= 100; b++) {
if (a < b && a * a + b * b == n * n) {
printf("(%d, %d, %d)\n", a, b, n);
impossible = 0;
}
}
}
if (impossible == 1) printf("impossible \n");
return 0;
#包括
int-power(int-base,int-power);
int main(){
int N;
printf(“输入数值:”);
scanf(“%d”和“&N”);
INTA、b、c;
对于(a=0;a
谢谢你回答我的问题。但如果我按照你告诉我的那样做,“不可能”一词将输出n 100次。。
#include <stdio.h>
int power(int base, int power);
int main(){
int N;
printf("INput the Num: ");
scanf("%d", &N);
int a, b, c;
for(a = 0; a < N ; a++) {
for(b = 0; b< N; b++) {
if ((a < b) && (b < N - a - b)) {
if (power(a, 2) + power(b, 2) == power(N - a - b, 2)) {
printf("%d^2 + %d^2 = %d^2 \n", a, b, N-a-b);
}
}
}
}
}
int power(int base, int power) {
int result = 1;
for(int i = 0; i < power ; i++) {
result *= base;
}
return result;
}