表現形式
要 驗證四方定理，因此我們可以使用窮舉法，對一個數，先用1一個數的平方和，2個數平方和，3個平方數的平方和，4個數的平方和進行驗證。如果成功返回1，失敗返回0；當窮舉完所有情況後，就可以驗證四方定理了，代碼如下：
#include <stdio.h>
#include <math.h>
int mode_1(int n)
{
if((int)sqrt(n) * (int)sqrt(n) == n){
printf("%d * %d = %d \n",(int)sqrt(n),(int)sqrt(n),n);
return 1;
}
else{
return 0;
}
}
int mode_2(int n)
{
int x,y;
for(x=1; x<sqrt(n); x++)
for(y=x; y<sqrt(n); y++){
if(x*x + y*y == n){
printf("%d^2 + %d^2 = %d\n",x,y,n);
return 1;
}
}
return 0;
}
int mode_3(int n)
{
int x,y,z;
for(x=1; x<sqrt(n); x++)
for(y=x; y<sqrt(n); y++)
for(z=y; z<sqrt(n); z++){
if(x*x + y*y + z*z == n){
printf("%d^2 + %d^2 + %d^2 = %d\n",x,y,z,n);
return 1;
}
}
return 0;
}
int mode_4(int n)
{
int x,y,z,t;
for(x=1; x<sqrt(n); x++)
for(y=x; y<sqrt(n); y++)
for(z=y; z<sqrt(n); z++)
for(t=z; t<sqrt(n); t++)
if(x*x + y*y + z*z + t*t == n){
printf("%d^2 + %d^2 + %d^2 + %d^2 = %d\n",x,y,z,t,n);
return 1;
}
return 0;
}
void proveFourSquares(int n)
{
if(mode_1(n))
printf("it has verified four squares\n");
else if(mode_2(n))
printf("it has verified four squares\n");
else if(mode_3(n))
printf("it has verified four squares\n");
else if(mode_4(n))
printf("it has verified four squares\n");
}
int main(int argc, char *argv[])
{
int n;
printf("please input a natural number\n");
scanf("%d",&n);
printf("--------------------------\n");
proveFourSquares(n);
return 0;
}
執行結果
如下：
peng@ubuntu:~/src/test/c/suanfa/miaoqu$ gcc 6.5.c -lm
peng@ubuntu:~/src/test/c/suanfa/miaoqu$ ./a.out
please input a natural number
120
--------------------------
2^2 + 4^2 + 10^2 = 120
it has verified four squares
peng@ubuntu:~/src/test/c/suanfa/miaoqu$ ./a.out
please input a natural number
10
--------------------------
1^2 + 3^2 = 10
it has verified four squares
