HDU ~ 2199 ~ Can you solve this equation?(二分+精度)
2017-11-30 08:18
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题意:给你一个Y,问有没有一个x且0<=x<=100使得8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,保留四位小数输出
思路:很容易看出函数在[0,100]是单调递增的,所以Y < F(0) 和Y > F(100)这两种情况无解,有解的时候二分找X就可以
思路:很容易看出函数在[0,100]是单调递增的,所以Y < F(0) 和Y > F(100)这两种情况无解,有解的时候二分找X就可以
#include<bits/stdc++.h> using namespace std; const double eps = 1e-8; double Y; double F(double x)//返回等号左边的值 { return 8.0*pow(x,4) + 7.0*pow(x,3) + 2.0*pow(x,2) + 3.0*x + 6.0; } int main() { int T; scanf("%d", &T); while(T--) { scanf("%lf", &Y); if(Y < F(0) || Y > F(100)){ printf("No solution!\n"); continue; }//没答案的情况 double l = 0.0, r = 100.0, m = l + (r - l) / 2.0; while(r - l > eps) { if(F(m) < Y) l = m + eps; else if(F(m) > Y) r = m - eps; else break; m = l + (r - l) / 2.0; } printf("%.4lf\n",m); } return 0; } /* 2 100 -4 */
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