hdoj 1795 The least one 【简单题】

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最新推荐文章于 2023-09-21 10:05:08 发布

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本文探讨了一个RPG游戏中英雄选择最优策略的问题，即如何选择与怪物等级互质且最小的英雄来最大化战斗效率。通过输入怪物等级Q，程序输出能够击杀至少Q个怪物的最小英雄等级，实现战斗的最小伤害。

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The least one

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 418 Accepted Submission(s): 147

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

2
3
7

Sample Output

5
11

题意：求大于Q的且与Q互质的最小的数。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
bool judge(int x)
{
    for(int i = 2; i <= sqrt(x); i++)
        if(x % i == 0) return false;
    return true;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int Q;
        scanf("%d", &Q);
        for(int i = Q + 1; ; i++)
        {
            if(judge(i))
            {
                printf("%d\n", i);
                break;
            }
        }
    }
    return 0;
}