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题目：
Can you solve this equation? |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 915 Accepted Submission(s): 436 |
Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100; Now please try your lucky. |
Input The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10); |
Output For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100. |
Sample Input 2 100 -4 |
Sample Output 1.6152 No solution! |
Author Redow |
Recommend lcy |
题目分析：
简单题。是用二分法来求方程组的解。使用二分法来求方程组的解的前提是该方程具有单调性。
代码如下：
/*
* a.cpp
*
* Created on: 2015年2月16日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double delta = 1e-6;//1e10表示10^10
double function(double x){
return 8*pow(x,4.0) + 7*pow(x,3) + 2*pow(x,2) + 3*x + 6;
}
/**
* 用二分法来求解方程组的解
* y:方程组右边的数
*/
double work(double y){
double mid;
double left = 0;
double right = 100;
while(right - left > delta){//如果右边的索引还比左边的索引大
mid = (left+right)/2;//计算中间值
if(function(mid) < y){//如果y大于中间值...注意这种写法
left = mid+(1e-7);//移动左边的索引。需要注意的是这里使用的偏移量应该比上面的小
}else{//否则
right = mid-(1e-7);//移动右边的索引
}
}
return (left+right)/2;//返回最后的值
}
int main(){
int t;
scanf("%d",&t);
while(t--){
double y;
// scanf("%d",&y);
cin >> y;
if(function(0) <= y && y <= function(100)){
printf("%.4lf\n",work(y));
// printf("%.4lf\n",find(y));
}else{
printf("No solution!\n");
}
}
return 0;
}
(hdu step 4.1.1)Can you solve this equation?(使用二分法来求解方程组的解)
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原文地址：http://blog.csdn.net/hjd_love_zzt/article/details/43851437
