HDU 2199 Can you solve this equation? 二分水题

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Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17700 Accepted Submission(s): 7873

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

2100-4

Sample Output

1.6152No solution!

Author

Redow

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题意很明确，就是给你一个y，求出来对应的x，结果保留4位小数，本来想用暴力直接搜，但头介绍用二分法优化一下，可以避免超时，就没直接暴力搜索，二分法也很简单，答案肯定在0-100之间（题意给的），所以没有答案的我这里返回-1.

ac代码

<span style="font-size:18px;"><strong>#include <stdio.h>int jdg(double head,double tail){if(head>=tail)return 1;return 0;}double result(double x){return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}double bin(double head,double tail,double value){double mid;if(value<6||value>807020306)return -1;mid=(head+tail)/2;while(!jdg(head,tail)){//防止精度过大，if((-1e-5)<(result(mid)-value)&&(result(mid)-value)<(1e-5))return mid;else if(result(mid)<value)head=mid;elsetail=mid;mid=(head+tail)/2; }return mid;}int main(){int t;double y,ans;scanf("%d",&t);while(t--){scanf("%lf",&y);ans=bin(0,100,y);if(ans<0)printf("No solution!\n");elseprintf("%.4lf\n",ans);}return 0;}</strong></span>

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