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
Recommend
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 <iostream> 3 using namespace std; 4 int i, j, sieve[10010]; 5 void is_prime() 6 { 7 for(i = 2; i < 10010; i++) 8 { 9 if(sieve[i] == 0) 10 for(j = i*i; j < 10010; j+=i) 11 sieve[j] = 1; 12 } 13 } 14 int main() 15 { 16 int m, t; 17 is_prime(); 18 scanf("%d", &t); 19 while(t--) 20 { 21 scanf("%d", &m); 22 for(i = m + 1; i < 10010; i++) 23 if(sieve[i] == 0) 24 break; 25 printf("%d ", i); 26 } 27 return 0; 28 }
1 /*孪生素数： 所谓孪生素数指的是间隔为 2 的相邻素数，它们之间的距离已经近得不能再近了。 2 3 4 5 6 若n≥6且n-1和n+1为孪生素数，那么n一定是6的倍数。 7 证明： 8 ∵ n-1和n+1是素数 ┈┈┈┈┈ ① 9 ∴ n-1和n+1是奇数 10 ∴ n是偶数，即n是2的倍数 ┈┈┈┈┈ ② 11 假设n不是3的倍数，得： 12 n=3x+1 或 n=3x+2， 13 如果n=3x+1，则n-1=3x，与①违背，故n≠3x+1； 14 如果n=3x+2，则n+1=3(x+1)，与①违背，故n≠3x+2； 15 ∴假设不成立，即n是3的倍数，又有②得结论： 16 n是6的倍数。 17 18 19 由上面的规律可以推出下面结论： 20 若x≧1且n=6x-1或n=6x+1不是素数，那么n一定不是2和3的倍数。 21 证明： 22 ∵n=6x-1或n=6x+1，即n=2(3x)-1或n=2(3x)+1或n=3(2x)-1或n=3(2x)+1。 23 ∴n一定不是2和3的倍数。 24 25 26 素数出现规律： 27 当n≧5时，如果n为素数，那么n mod 6 = 1 或 n mod 6 = 5，即n一定出现在6x（x≥1）两侧。 28 证明： 29 当x≥1时，有如下表示方法： 30 ┈┈ 6x，6x+1，6x+2，6x+3，6x+4，6x+5，6(x+1），6(x+1)+1┈┈ 31 不在6x两侧的数为6x+2，6x+3，6x+4，即2(3x+1)，3(2x+1)，2(3x+2)，它们一定不是素数，所以素数一定出现在6x的两侧。*/ 32 bool isPrime(int num) 33 { 34 if (num == 2 || num == 3) 35 { 36 return true; 37 } 38 if (num % 6 != 1 && num % 6 != 5) 39 { 40 return false; 41 } 42 for (int i = 5; i*i <= num; i += 6) 43 { 44 if (num % i == 0 || num % (i+2) == 0) 45 { 46 return false; 47 } 48 } 49 return true; 50 }