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; |
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; }
时间: 2024-10-06 12:07:21