bzoj 1041[HAOI2008]圆上的整点 - 数论

1041: [HAOI2008]圆上的整点

Time Limit: 10 Sec  Memory Limit: 162 MB

Description

求一个给定的圆(x^2+y^2=r^2)，在圆周上有多少个点的坐标是整数。

Input

只有一个正整数n,n<=2000 000 000

Output

整点个数

Sample Input

4

Sample Output

4

$ x ^ 2 + y ^ 2 = r ^2$

移项

$x ^ 2 = r ^ 2 - y ^ 2$

$x ^ 2 = (r - y) * (r + y)$

 令  d = gcd(r - y, r + y)

设 $A = \frac{r - y}{d} = a ^ 2$, $B = \frac{r + y}{d} = b ^ 2$ (gcd(A , B) == 1)

则 $a ^ 2 + b ^ 2 = \frac{r * 2}{d}$

所以d 为 2 * r 的约数

再枚举 a = 1 ~ sqrt($\frac{d}{2}$) 即可

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long

using namespace std;

LL ans = 0;
LL r;
inline LL read()
{
    LL x = 0, w = 1; char ch = 0;
    while(ch < '0' || ch > '9') {
        if(ch == '-') {
            w = -1;
        }
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * w;
}

int gcd(int a, int b)
{
    if(b == 0) {
        return a;
    } else {
        return gcd(b, a % b);
    }
}
int main()
{
    r = read();
    for(int d = 1; d <= sqrt(2 * r); d++) {
        if((2 * r) % d == 0) {
            for(int a = 1; a < sqrt(2 * r / (2 * d)); a++) {
                int b = (int)sqrt(2 * r / d - a * a);
                if(b == sqrt(2 * r / d - a * a) && gcd(b, a) == 1) {
                //    cout<<a<<" "<<b<<endl;
                    ans++;
                }
            }
            if(d != (2 * r) / d) {
                for(int a = 1; a < sqrt(d / 2); a++) {
                    int b = (int)sqrt(d - a * a);
                    if(b == sqrt(d - a * a) && gcd(b, a) == 1) {
                //        cout<<a<<" "<<b<<endl;
                        ans++;
                    }
                }
            }    
        }
    }
    ans = ans * 4 + 4;
    printf("%lld\n", ans);
    return 0;
}