ZOJ 1505 Solitaire

杜仲当归 — Tue, 26 Feb 2008 06:20:00 GMT

摘要: 2761550 2008-02-26 14:10:05 Accepted 1505 C++ 00:00.11 3480K C.D.=.= ... 阅读全文

杜仲当归 2008-02-26 14:20 发表评论

ZOJ 2330 A^B == B^A?

杜仲当归 — Mon, 11 Feb 2008 06:18:00 GMT

菜题。即便是这样我还是写了三遍……步长开的太大，结果精度不高。

把A^B == B^A两边求对数，很容易得到ln(x)/x的函数。求导，此函数在e两侧单调。故可用二分。

1/*
2    二分逼近ln(x)/x = ln(a)/a
3*/
4
5#include <cstdio>
6#include <cmath>
7
8#define ZERO 1e-14
9#define E exp(1.0f)
10
11double A;
12
13int main ()
14{
15    //freopen ( "in.txt", "r", stdin );
16    while ( scanf ( "%lf", &A ) != EOF )
17    {
18        if ( A - E > ZERO )
19            printf ( "-1\n" );
20        else
21        {
22            double high, low, mid, d;
23            high = 45, low = E;
24            while ( high - low >= ZERO )
25            {
26                mid = ( high + low ) / 2;
27                d = log ( mid ) / mid - log ( A ) / A;
28                if ( d < ZERO && d > -ZERO )
29                    break;
30                else if ( d > ZERO )
31                {
32                    low = mid + 0.0000001;
33                }
34                else if ( d < -ZERO )
35                {
36                    high = mid - 0.0000001;
37                }
38            }
39            printf ( "%0.5f\n", mid );
40            d = fabs ( log ( mid ) / mid - log ( A ) / A );
41            //printf ( "error %0.8f\n", d );
42        }
43    }
44    return 0;
45}

杜仲当归 2008-02-11 14:18 发表评论

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ZOJ 1505 Solitaire

ZOJ 2330 A^B == B^A?