1041: [HAOI2008]圆上的整点
Time Limit: 10 Sec Memory Limit: 162 MBDescription
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #define LL long long 7 8 using namespace std; 9 10 LL ans = 0; 11 LL r; 12 inline LL read() 13 { 14 LL x = 0, w = 1; char ch = 0; 15 while(ch < '0' || ch > '9') { 16 if(ch == '-') { 17 w = -1; 18 } 19 ch = getchar(); 20 } 21 while(ch >= '0' && ch <= '9') { 22 x = x * 10 + ch - '0'; 23 ch = getchar(); 24 } 25 return x * w; 26 } 27 28 int gcd(int a, int b) 29 { 30 if(b == 0) { 31 return a; 32 } else { 33 return gcd(b, a % b); 34 } 35 } 36 int main() 37 { 38 r = read(); 39 for(int d = 1; d <= sqrt(2 * r); d++) { 40 if((2 * r) % d == 0) { 41 for(int a = 1; a < sqrt(2 * r / (2 * d)); a++) { 42 int b = (int)sqrt(2 * r / d - a * a); 43 if(b == sqrt(2 * r / d - a * a) && gcd(b, a) == 1) { 44 // cout<<a<<" "<<b<<endl; 45 ans++; 46 } 47 } 48 if(d != (2 * r) / d) { 49 for(int a = 1; a < sqrt(d / 2); a++) { 50 int b = (int)sqrt(d - a * a); 51 if(b == sqrt(d - a * a) && gcd(b, a) == 1) { 52 // cout<<a<<" "<<b<<endl; 53 ans++; 54 } 55 } 56 } 57 } 58 } 59 ans = ans * 4 + 4; 60 printf("%lld ", ans); 61 return 0; 62 }