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  • poj 1183

    题目:http://poj.org/problem?id=1183

    前面给了好长的一段描述,最后面的那个公式是关键:arctan(p)+arctan(q)=arctan[(p+q)/(1-pq)],然后题目是给出 a 求 (b + c)满足:arctan(1/a)=arctan(1/b)+arctan(1/c) ,根据掐面的公式可以化简得 : a = (b * c - 1) / (b + c)。令 b = (a + m)   c = (a + n)然后代入化简得 m * n = a ^ 2 + 1;所以枚举 m 或 n 然后求出另一个就行了

    View Code
     1 int main()
     2 {
     3     int a,b,c;
     4     ll i,j;
     5     while(~scanf("%d",&a))
     6     {
     7         int tem = a;
     8         ll kemp = (ll) (a) * (ll) (a) + 1;
     9         for(i = tem + 1; i >= 1; i--)
    10         {
    11             if(kemp % i == 0)
    12             {
    13                 j = kemp / i;
    14                 break;
    15             }
    16         }
    17         cout<<(2 * tem + i + j)<<endl;
    18     }
    19     return 0;
    20 }
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  • 原文地址:https://www.cnblogs.com/fxh19911107/p/2779600.html
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