Problem F
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= 0where0 <=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,tandu(where0 <=p,r<= 20and-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 ofx, correct upto 4 decimal places, or the string "No solution", whichever is applicable.
Sample Input0 0 0 0 -2 11 0 0 0 -1 21 -1 1 -1 -1 1Sample Output0.7071No solution0.7554
题意：给出p,q,r,s,t,u的值，根据方程式求出方程的近似解。
思路：因为方程是递减的，对于两个端点，如果f(0)<0或者是f(1)>0 ,则说明方程无解。取两个端点的中点mid 如果f(0)*f(mid)<0 ,说明方程的解在[0,mid]中，否则解在[mid,1]中，这样不断进行下去，，区间会收敛于一点，即为最后的近似解
#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;#define eps 1e-8double 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;//exp(x) 指的是e的x次方。}int main(){double left,right,mid;while(~scanf("%lf %lf %lf %lf %lf %lf",&p,&q,&r,&s,&t,&u)) {if(Solve(0)<0||Solve(1)>0)printf("No solution\n");else {left=0;right=1;while((right-left)>=eps) {mid=(left+right)/2;if(Solve(mid)<0)right=mid;elseleft=mid;}printf("%.4lf\n",mid);}}return 0;}
缘是浪漫的相遇，瞬间让你我的心化为永恒！
