Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16850    Accepted Submission(s): 7491


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!  
  题目给你一个y，让你判断它的解是否在0~100内，开始没想到做法啊，其实做过了，还是不太熟悉二分 二分代码： /* 以为可以用高中的数学公式推出来呢,看样子是不能了，开始想到二分， 但是对小数操作不怎么会，就弃了，看了别人写的觉得就是自己不熟练二分啊 */ #include #include #include #include #include using namespace std; const double zero = 1e-6; double solve(double x) { return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6; } int main() { int t; cin >> t; double y; while(t--) { scanf("%lf",&y); if(solve(0)>y || y>solve(100)) { printf("No solution! "); } else { //用二分的方法，可谓精妙 double l = 0.0,r = 100.0,mid = (l+r)/2; while(fabs(solve(mid) - y) > zero)//”相等“ { if(solve(mid)>y) r = mid - 1; else l = mid + 1; mid = (l+r)/2; } printf("%.4f ",mid); } } return 0; }   因为是求一个高次方程的解，大部分高次方程都不存在求根公式，用牛顿迭代法可以求解的近似值，但是每次迭代的结果都不是精确值。此处又要扯上极限了，连续函数迭代求得的解是会收敛于零点的，所以最终结果就极其接近真实解了。具体解释见牛顿迭代法
公式          
牛顿迭代法代码： #include #include #include #include #include using namespace std; const double zero = 1e-6; double y; double solve(double x) { return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6; } double der(double x)//上式求导 { return 32*x*x*x + 21*x*x + 4*x + 3; } double NM(double x) { int cnt = 1; while(fabs(solve(x) - y) > zero) { x = x - (solve(x)-y)/der(x);//牛顿迭代法 if(++cnt > 30) return 0;//break; } return x; } int main() { int t; cin >> t; while(t--) { scanf("%lf",&y); bool flag = false; double k; for(double i = 0.0;i < 100;i++) { k = NM(i); if(k && k >= 0.0 && k <= 100.0) { flag = true; break; } } if(flag) printf("%.4f ",k); else printf("No solution! "); } return 0; } 代码参考自静涛,退出方式待研究