HDOJ 题目1795 The least one（水题）

最新推荐文章于 2020-04-17 15:02:57 发布

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本文介绍了一个游戏编程问题，玩家需要选择英雄角色击败怪物。每个英雄都有一个受尊敬的值，能够击败那些受尊敬值小于自己且互质的怪物。任务是找到最小受尊敬值的英雄，以完成杀死指定数量怪物的目标。

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The least one

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 376 Accepted Submission(s): 132

Problem Description

In the RPG game “go back ice age”(I decide to develop the game after my undergraduate education), all heros have their own respected value, and the skill of killing monsters is defined as the following rule: one hero can kill the monstrers whose respected values are smaller then himself and the two respected values has none common factor but 1, so the skill is the same as the number of the monsters he can kill. Now each kind of value of the monsters come. And your hero have to kill at least M ones. To minimize the damage of the battle, you should dispatch a hero with minimal respected value. Which hero will you dispatch ? There are Q battles, in each battle, for i from 1 to Q, and your hero should kill Mi ones at least. You have all kind of heros with different respected values, and the values(heros’ and monsters’) are positive.

Input

The first line has one integer Q, then Q lines follow. In the Q lines there is an integer Mi, 0<Q<=1000000, 0<Mi<=10000.

Output

For each case, there are Q results, in each result, you should output the value of the hero you will dispatch to complete the task.

Sample Input

Sample Output

  
   5
11

Author

wangye

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

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wangye | We have carefully selected several similar problems for you: 1793 1796 1797 1794 1788

题目说的好抽象啊，，就是找第一个比n大的素数，，原本想用不用二分找啊，，，居然不用也过了，，水

ac代码

#include<stdio.h>
#include<string.h>
int a[1000100],k;
void fun()
{
	int i,j;
	k=0;
	for(i=2;i<1000100;i++)
	{
		int w=1;
		for(j=2;j*j<=i;j++)
			if(i%j==0)
			{
				w=0;
				break;
			}
		if(w)
			a[k++]=i;
	}
}
int main()
{
	int t,i;
	fun();
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		for(i=0;i<k;i++)
		{
			if(a[i]>n)
				break;
		}
		printf("%d\n",a[i]);
	}
}