1267: 金币

本文介绍了一种特殊的金币发放模式,骑士的金币数量随时间增加而变化的计算方法。通过使用双层循环的C++代码实现,详细解析了如何计算前K天骑士获得的金币总数。

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1267: 金币

时间限制: 1 Sec  内存限制: 128 MB
提交: 45  解决: 27
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命题人:admin

题目描述

国王将金币作为工资,发放给忠诚的骑士。第一天,骑士收到一枚金币;之后两天 (第二天和第三天),每天收到两枚金币;之后三天(第四、五、六天),每天收到三枚 金币;之后四天(第七、八、九、十天),每天收到四枚金币……;这种工资发放模式 会一直这样延续下去:当连续 N 天每天收到 N 枚金币后,骑士会在之后的连续 N+1 天 里,每天收到 N+1 枚金币。 请计算在前 K 天里,骑士一共获得了多少金币。 

 

输入

输入文件只有 1 行,包含一个正整数 K,表示发放金币的天数。 

 

输出

输出文件只有 1 行,包含一个正整数,即骑士收到的金币数。 

 

样例输入

6

 

样例输出

14

 

提示

【输入输出样例  说明】 
骑士第一天收到一枚金币;第二天和第三天,每天收到两枚金币;第四、五、六天, 每天收到三枚金币。因此一共收到 1+2+2+3+3+3=14 枚金币。 


【数据说明】  对于 100%的数据,1 ≤ K ≤ 10,000。 

 

整个程序的关键部分就是让以后的i天加上相应的i次,这样的话用两个嵌套的for循环组合起来,外侧循环标记之后几天,内层循环标记之后i天要加上相应的i次,详细题解请看代码

#include<iostream>
#include<string>
using namespace std;

int main()
{

	int k;
	cin >> k;//k记录前k天里
	long long int sum = 0;//用来记录总共的金币数
	int temp;
	int count = 0;//变量count用来统计sum加的次数
	bool flag = false;
	for (int i = 1; i <= k; i++) {//外层循环中的变量i代表题目中的“之后几天”
		for (int j = 1; j <= i; j++) {/*内层循环从1开始到i结束,使外层循环中的变量i加上相应的i次,
									  对应题目中的“之后i天,每天都收到相应的i个金币,比如i等于3时,
									  内层循环3次,就会使sum加上3个3;
		 	                           ”*/
			count++;
			if (count > k) {//当count大于k天时就跳出循环
				break;
				flag = true;//flag标为true,以便使外层循环结束
			}
			sum += i;//内外两层循坏合起来,就可以让i加上相应的i次
			
		}
		if (flag)
			break;
	}

	cout << sum << endl;
	return 0;
}


 

 

分类

 

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