The least one

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 534    Accepted Submission(s):
203

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

Author

wangye

Source

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

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题意：杀怪兽游戏，英雄的能力要大于怪兽；并且两者能力都为质数，综合可得所求答案为 大于给出数的最小素数；

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 int sieve[10010];
 6 void is_prime()
 7 {
 8     memset(sieve, 0, sizeof(sieve));
 9     for(int i = 2; i<10010; i++)
10     {
11         if(sieve[i] == 0)
12         {
13             for(int j = 2 * i; j<10010; j+=i)
14                 sieve[j]=1;
15         }
16     }
17 }
18 int main()
19 {
20     int m, t, i;
21     scanf("%d", &t);
22     is_prime();
23     while(t--)
24     {
25         scanf("%d", &m);
26         for(i=m+1; i<10010; i++)
27             if(sieve[i]==0)
28                 break;
29         printf("%d\n", i);
30     }
31     return 0;
32 }