題目連結：http://poj.org/problem?id=3122
題目大意：
有 n 塊披薩（大小不一樣）， f 個人分，包含主人自己 f+1 人；
每人吃的披薩必須是一塊披薩上切下來的。每個人吃的披薩相等，披薩可以有剩餘。求每人吃的最大披薩面積。
思路：
- 假設每人分得的披薩面積等效為半徑 R的圓；
- 每塊披薩可以分給幾個人呢？ r[i] 表示披薩半徑，則是 r[i]2/R2 取整個人
- 然後全部累加起來，如果總和大于等于 f+1，則每個人還有分更大的披薩的可能，R取值增大
- 如果總和小于 f+1，則每個人分的太大了，不夠分的，R取值減小
- R的取值範圍在（0，max（r[i]））
Wrong Answer代碼
/**
* @description: 有 n 塊披薩（大小不一樣）， f 個人分，包含主人自己 f+1 人；
* 每人吃的披薩必須是一塊披薩上切下來的。求
* @author: michael ming
* @date: 2019/4/20 0:23
* @modified by:
*/
#include <iostream>
#include <iomanip>
#define PI 3.14159265359
using 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;
} 複制
AC代碼（主要修改，精度問題，把求人數的除法改成減法）
/**
* @description: 有 n 塊披薩（大小不一樣）， f 個人分，包含主人自己 f+1 人；
* 每人吃的披薩必須是一塊披薩上切下來的。求
* @author: michael ming
* @date: 2019/4/20 0:23
* @modified by:
*/
#include <iostream>
#include <iomanip>
#include <math.h>
#include <algorithm>
#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; //改成減法，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;
} 複制