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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12766 Accepted Submission(s):
5696
#include<stdio.h> #include<string.h> double f(double v)//用来计算方程结果(即y的值) { return 8*v*v*v*v+7*v*v*v+2*v*v+3*v+6; } int main() { int t; double y,l,r,mid; scanf("%d",&t); while(t--) { scanf("%lf",&y); if(y<f(0)||y>f(100)) { printf("No solution!\n"); continue; } l=0;r=100;mid=0; while((r-l) > 1e-10)//题目要求精度为小数点后四位,最好把精度调高 { mid=(l+r)/2;//每次折半取x的范围 if(f(mid) < y) l=mid; else r=mid; } printf("%.4lf\n",mid); } }
hdoj 2199 Can you solve this equation?【浮点型数据二分】
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原文地址:http://www.cnblogs.com/tonghao/p/4688714.html