标签：iostream eal please 情况 cin mes while printf tween
InputThe 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);OutputFor 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!
这种题一般都很注重精度的，可以拿两个端点来做验证，6的时候应该为0.0000,807020306的时候应该为100.0000
其中while(ri-le<=0.000000001)这里的0尽量多一点，而判断的时候的0的个数要看情况，有些当满足
ri-le<=0.000000001时，说不定判断还不满足。
#include <iostream> #include <stack> #include <string.h> #include <stdio.h> #include<queue> #include<algorithm> #define ll long long using namespace std; int main() { int t; cin>>t; while(t--) { double y; cin>>y; double le,ri,mid; le=0;ri=100; bool f=0; while(ri-le>=0.00000000000001) { mid=(le+ri)/2; //cout<<8*mid*mid*mid*mid+7*mid*mid*mid+2*mid*mid+3*mid+6-y<<endl; if(8*mid*mid*mid*mid+7*mid*mid*mid+2*mid*mid+3*mid+6-y<0.00001&&8*mid*mid*mid*mid+7*mid*mid*mid+2*mid*mid+3*mid+6-y>=0) { f=1; break; } else if(8*mid*mid*mid*mid+7*mid*mid*mid+2*mid*mid+3*mid+6-y<0) { le=mid; } else ri=mid; } if(f) printf("%.4lf\n",mid); else printf("No solution!\n"); } return 0; }
HDU 2199 Can you solve this equation?(二分精度）
标签：iostream eal please 情况 cin mes while printf tween
原文地址：https://www.cnblogs.com/caiyishuai/p/9541313.html
