题目链接如下：

Online Judge

这道题目我想到了打表做，但是没想到用log来简化……写了一个比较复杂的版本，严重超时。后来看了别人的题解才做出来。UVa 11809 Floating-Point Numbers（浮点数）_ShannonNansen的博客-CSDN博客

我的代码如下：

#include <iostream>
#include <string>
#include <cmath>
const double EPS = 0.0001;std::string str;
int E, temp;
double M, t, a;
double ans1[10][31];
int ans2[10][31];int main(){for(int i = 0; i <= 9; ++i){M = 1 - 1.0 / (1 << (i + 1));for(int j = 1; j <= 30; ++j){E = (1 << j) - 1;t = log10(M) + E * log10(2);ans2[i][j] = t;ans1[i][j] = pow(10, t - ans2[i][j]);}}while(std::cin >> str && str != "0e0"){a = std::stod(str.substr(0, 17));temp = std::stoi(str.substr(18));for(int i = 0; i <= 9; ++i){for(int j = 1; j <= 30; ++j){if(ans2[i][j] == temp && fabs(a - ans1[i][j]) < EPS){printf("%d %d\n", i, j);i = 10;break;}}}}return 0;
}

我严重超时的版本如下：

#include <iostream>
#include <string>
#include <cmath>std::string str;
int M, E, t, temp, power, prevpower;
double a, tempa, b, k, preva;
double ans1[10][31];
int ans2[10][31];int main(){a = 0;k = 1;for(M = 0; M <= 9; ++M){k *= 0.5;a += k;t = 0;for(E = 1; E <= 30; ++E){temp = t + 1;t = 2 * t + 1;tempa = E == 1 ? a : preva;power = E == 1 ? 0 : prevpower;while(temp){while(tempa < 10 && temp){tempa *= 2;temp--;}if(tempa >= 10){tempa /= 10;power++;}}preva = tempa;prevpower = power;ans1[M][E] = tempa;ans2[M][E] = power;}}while(std::cin >> str && str != "0e0"){a = std::stod(str.substr(0, 17));temp = std::stoi(str.substr(18));for(M = 0; M <= 9; ++M){for(E = 1; E <= 30; ++E){if(ans2[M][E] == temp && fabs(a - ans1[M][E]) < 0.0001){printf("%d %d\n", M, E);M = 10;break;}}}}return 0;
}