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题目链接：

题目大意：

有 n 块披萨（大小不一样）， f 个人分，包含主人自己 f+1 人；

思路：

• 假设每人分得的披萨面积等效为半径 R的圆；
• 每块披萨可以分给几个人呢？ r[i] 表示披萨半径，则是 r[i]²/R² 取整个人
• 然后全部累加起来，如果总和大于等于 f+1，则每个人还有分更大的披萨的可能，R取值增大
• 如果总和小于 f+1，则每个人分的太大了，不够分的，R取值减小
• R的取值范围在（0，max（r[i]））

Wrong Answer代码

/** * @description: 有 n 块披萨（大小不一样）， f 个人分，包含主人自己 f+1 人； *                  每人吃的披萨必须是一块披萨上切下来的。求 * @author: michael ming * @date: 2019/4/20 0:23 * @modified by:  */#include 
   
    #include 
    
     #define PI 3.14159265359using namespace std;const double error = 1e-7;double find_max_R(size_t pizza_num, int *r_pizza, double r_low, double r_high, size_t people){
       double R_we_want = r_low+(r_high-r_low)/2;    size_t people_get_pizza = 0;    while(r_high - r_low > error)    {
           people_get_pizza = 0;        R_we_want = r_low+(r_high-r_low)/2;        for(int i = 0; i < pizza_num; ++i)            people_get_pizza += (int)(r_pizza[i]*r_pizza[i]/(R_we_want*R_we_want));        if(people_get_pizza >= people)            r_low = R_we_want;        else            r_high = R_we_want;    }    return R_we_want;}int main(){
       size_t t, pizza_num, friend_num;    double r_max_pizza = 0;    cin >> t;    while(t--)    {
           cin >> pizza_num >> friend_num;        int *r_pizza = new int [pizza_num];        for(int i = 0; i < pizza_num; ++i)        {
               cin >> r_pizza[i];            r_max_pizza = r_pizza[i] > r_max_pizza ? r_pizza[i] : r_max_pizza;        }        r_max_pizza = find_max_R(pizza_num,r_pizza,0,r_max_pizza,friend_num+1);        cout << setiosflags(ios::fixed) << setprecision(4) << PI*r_max_pizza*r_max_pizza << endl;        delete[] r_pizza;        r_pizza = NULL;    }    return 0;}
    
   

AC代码（主要修改，精度问题，把求人数的除法改成减法）

/** * @description: 有 n 块披萨（大小不一样）， f 个人分，包含主人自己 f+1 人； *                  每人吃的披萨必须是一块披萨上切下来的。求 * @author: michael ming * @date: 2019/4/20 0:23 * @modified by:  */#include 
   
    #include 
    
     #include 
     
      #include 
      
       #define PI acos(-1.0)using namespace std;const double error = 1e-7;double find_max_R(size_t pizza_num, double *s_pizza, double s_low, double s_high, size_t people){
       double S_we_want = s_low+(s_high-s_low)/2.0;    size_t people_get_pizza = 0;    while(s_high - s_low > error)    {
           people_get_pizza = 0;        S_we_want = s_low+(s_high-s_low)/2.0;        for(int i = 0; i < pizza_num; ++i)        {
               double temp = s_pizza[i];            while(temp-S_we_want>=0)            {
                   temp -= S_we_want;	//改成减法，不易丢失精度                people_get_pizza++;            }        }        if(people_get_pizza >= people)            s_low = S_we_want;        else            s_high = S_we_want;    }    return S_we_want;}int main(){
       size_t t, pizza_num, friend_num;    double s_max_pizza = 0.0;    cin >> t;    while(t--)    {
           cin >> pizza_num >> friend_num;        double *s_pizza = new double [pizza_num];        for(int i = 0; i < pizza_num; ++i)        {
               cin >> s_pizza[i];            s_pizza[i] *= s_pizza[i];        }        sort(s_pizza, s_pizza+pizza_num);        s_max_pizza = find_max_R(pizza_num,s_pizza,0,s_pizza[pizza_num-1],friend_num+1);        cout << setiosflags(ios::fixed) << setprecision(4) << PI*s_max_pizza << endl;        delete[] s_pizza;        s_pizza = NULL;    }    return 0;}
      
     
    
   

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