UVA 10341 - Solve It（二分）

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

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二分查找方程的解，注意边界也要判断

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

double p,q,r,s,t,u;

double solve(double x)
{
    return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;
}

int main()
{
    while(~scanf("%lf%lf%lf%lf%lf%lf",&p,&q,&r,&s,&t,&u))
    {
        double left=0,right=1,mid;
        if(solve(left)*solve(right)>0)//如果值均为正或均为负，则01之间没有解
        {
            printf("No solution\n");
            continue;
        }
        if(solve(left)==0)
        {
            printf("0.0000\n");
            continue;
        }
        if(solve(right)==0)
        {
            printf("1.0000\n");
            continue;
        }
        while(left<=right)
        {
            mid=(left+right)/2;
            if(solve(mid)>-(1e-9)&&solve(mid)<1e-9) break;
            if(solve(left)*solve(mid)<0) right=mid;
            else left=mid;
        }
        printf("%.4lf\n",mid);
    }
    return 0;
}