Solve It - UVa 10341 二分

最新推荐文章于 2020-07-27 19:43:35 发布

原创最新推荐文章于 2020-07-27 19:43:35 发布 · 542 阅读

0 ·

CC 4.0 BY-SA版权

UVa 同时被 2 个专栏收录

218 篇文章

订阅专栏

数据结构

67 篇文章

订阅专栏

本文介绍了一种通过二分法求解特定形式方程在区间[0,1]内的根的方法，并提供了一段AC代码实现。针对方程p*e^-x + q*sin(x) + r*cos(x) + s*tan(x) + t*x^2 + u = 0，讨论了精度设置与输出结果的关系。

Solve It

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

Solve the equation:
p*e^-x+ q*sin(x) + r*cos(x) + s*tan(x) + t*x² + u = 0
where 0 <= x <= 1.

Input

Input consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers in a single line: p, q, r, s, t and u(where 0 <= p,r <= 20 and -20 <= q,s,t <= 0). There will be maximum 2100 lines in the input file.

Output

For each set of input, there should be a line containing the value of x, correct upto 4 decimal places, or the string "No solution", whichever is applicable.

Sample Input

0 0 0 0 -2 1

1 0 0 0 -1 2

1 -1 1 -1 -1 1

Sample Output

0.7071

No solution

0.7554

题意：求上式中0的解x。

思路：二分求解。有一点我很奇怪的是，当eps=1e-6时WA，至少要1e-7，但是答案只需要输出前4位，这点我很无语，求解释。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
double p,q,r,s,t,u,eps=1e-7;
double solve(double x)
{
    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
}
int main()
{
    int n,i,j,k;
    double L,R,mi;
    while(~scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u))
    {
        if(solve(0)*solve(1)>0)
        {
            printf("No solution\n");
            continue;
        }
        L=0;R=1;
        while(R-L>eps)
        {
            mi=(L+R)/2;
            if(solve(mi)*solve(L)>0)
              L=mi;
            else
              R=mi;
        }
        printf("%.4f\n",L);
    }
}