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  • bzoj 1041 圆上的整点 分类: Brush Mode 2014-11-11 20:15 80人阅读 评论(0) 收藏

    这里先只考虑x,y都大于0的情况

    如果x^2+y^2=r^2,则(r-x)(r+x)=y*y

    令d=gcd(r-x,r+x),r-x=d*u^2,r+x=d*v^2,显然有gcd(u,v)=1且u<v

    有2r=d*(u^2+v^2),y=d*u*v,x=d(v^2-u^2)/2

    枚举2r的约数d,再花费sqrt(2r/d)的时间枚举u,求出v=sqrt(2r/d-u^2)然后判断gcd(u,v)=1

    最后结果乘以4(四个象限)+4(坐标轴上)即可

    /**************************************************************
        Problem: 1041
        User: ******
        Language: C++
        Result: Accepted
        Time:92 ms
        Memory:1284 kb
    ****************************************************************/
     
    #include <vector>
    #include <list>
    #include <limits.h>
    #include <map>
    #include <set>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <bitset>
    #include <algorithm>
    #include <functional>
    #include <numeric>
    #include <utility>
    #include <sstream>
    #include <iostream>
    #include <iomanip>
    #include <cstdio>
    #include <cmath>
    #include <cstdlib>
    #include <ctime>
    #include <string.h>
    #include <stdlib.h>
    #include <cassert>
     
    using namespace std;
     
    int main() {
        long long r;
        cin >> r;
        long long d, a, b;
        long long ans = 0;
        for (d = 1; d * d <= 2*r; ++d) {
            if ((2 * r) % d) continue;
            long long t = (2 * r) / d;
            for (a = 1; a * a <= t; ++a) {
                b = sqrt(t - a * a);
                if (b * b != t - a * a) continue;
                if (a >= b) continue;
                if (__gcd(a, b) > 1) continue;
                // cout << d << " " << a << " " << b << endl;
                // cout << "X: " << - a * a * d + r << endl;
                ++ans;
            }
            if (d * d == 2 * r) continue;
            t = d;      
            for (a = 1; a * a <= t; ++a) {
                b = sqrt(t - a * a);
                if (b * b != t - a * a) continue;
                if (a >= b) continue;
                if (__gcd(a, b) > 1) continue;
                // cout << t << " " << a << " " << b << endl;
                // cout << "X: " << - a * a * (2*r/d) + r << endl;
                ++ans;
            }
        }
        ans = (ans + 1) * 4;
        cout << ans << endl;
        return 0;
    }
    
    转摘至:http://blog.csdn.net/lwfcgz/article/details/39927801

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  • 原文地址:https://www.cnblogs.com/you-well-day-fine/p/4671612.html
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