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简单的二分HDU2199
原题http://acm.hdu.edu.cn/showproblem.php?pid=2199
Can you solve this equation?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8441 Accepted Submission(s): 3902
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.
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
//一道简单的二分。首先判断下在那些范围是一定有解的 //然后不断的缩小范围就可以了 #include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <limits.h> #include <ctype.h> #include <string.h> #include <string> #include <math.h> #include <algorithm> #include <iostream> #include <stack> #include <queue> #include <deque> #include <vector> #include <set> #include <map> using namespace std; __int64 Y; const double d = 1e-8; double f(double m){ if(Y-(8*m*m*m*m+7*m*m*m+2*m*m+3*m+6)>=1e-20){ return 1; } else{ return 0; } } //d//ouble f1(double x){/ // return (8*x*x*x*x+7*x*x*x+2*x*x+3*x+6-Y); //} int main(){ int T; while(~scanf("%d",&T)){ while(T--){ scanf("%I64d",&Y); if(Y<6 || Y>807020306){ printf("No solution!\n"); continue; } else{ double l = 0.0; double r = 100.0; int flag = 0; while(r-l > d){ double mid = (l+r)/2; if(f(mid) == 1){ l = mid; } else{ r = mid; } } printf("%.4lf\n",l); //if(flag == 0){ // printf("No solution!\n"); //} } } } return 0; }
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