UVA 10341 二分搜索
Solve the equation:
p ∗ e
−x + q ∗ sin(x) + r ∗ cos(x) + s ∗ tan(x) + t ∗ x
2 + 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 up to 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:求解这个等式
题解:我们分析下这个等式,就发现是个单调递减的,二分[0,1]就好了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
//meek///#include<bits/stdc++.h> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include<iostream> #include<bitset> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int N = 100010; const int inf = 0x3f3f3f3f; const int MOD = 100003; const double eps = 0.000001; double q,p,r,s,t,u,a,b; double check(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 a = 0.0,b = 1.0; if(check(a) * check(b) > 0) { printf("No solution\n");continue; } else if(fabs(check(0.0)) <= eps){ printf("0.0000\n");continue; } else if(fabs(check(1.0)) <= eps) { printf("1.0000\n");continue; } if(check(0)>0) { a = 1.0; b = 0.0; } else { b = 1.0; a = 0.0; } while(1) { double x = (a+b)/2; double temp = check(x); if(fabs(temp) <= eps) { printf("%.4f\n",x); break; } else if(temp > 0) { b = x; } else a = x; } } return 0; }