使用C++计算力量P可以杀死的最大人数


给定任务是找到使用力量P可以杀死的最大人数。考虑一个包含无限人数的行,每个人都有从1开始的索引号。

第s个人的力量用s2表示。杀死一个具有s力量的人后,你的力量也会减少s。

让我们用一个例子来理解我们必须做什么:

输入

P = 20

输出

3

Explore our latest online courses and learn new skills at your own pace. Enroll and become a certified expert to boost your career.

解释

Strength of 1st person = 1 * 1 = 1 < 20, therefore 1st person can be killed.
Remaining strength = P – 1 = 20 – 1 = 19
Strength of 2nd person = 2 * 2 = 4 < 19, therefore 2nd person can be killed.
Remaining strength = P – 4 = 19 – 4 = 15
Strength of 3rd person = 3 * 3 = 9 < 15, therefore 3rd person can be killed.
Remaining strength = P – 9 = 15 – 9 = 6
Strength of 4th person = 4 * 4 = 16 > 6, therefore 4th person cannot be killed.
Output = 3

输入

30

输出

4

下面程序中使用的方法如下

  • 在main()函数中,初始化类型为int的P = 30,因为它将存储力量,并将其传递到Max()函数。

  • 在Max()函数中,初始化类型为int的s = 0和P = 0。

  • 从j = 1循环到j * j <= P

  • 设置s = s + (j * j),如果s <= P,则将ans加1,否则中断;

  • 返回ans。

示例

演示

#include <bits/stdc++.h>
using namespace std;
int Max(int P){
   int s = 0, ans = 0;
   for (int j = 1; j * j <= P; j++){
      s = s + (j * j);
      if (s <= P)
         ans++;
      else
         break;
   }
   return ans;
}
//main function
int main(){
   //Strength
   int P = 30;
   cout << Maximum number of people that can be killed with strength P are: ”<<Max(P);
   return 0;
}

输出

Maximum number of people that can be killed with strength P are: 4

更新于:2020年8月3日

153 次浏览

启动您的职业生涯

完成课程获得认证

开始
广告