HDU 1216 Assistance Required 埃拉托色尼色筛法

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1216

思路：色筛法

代码（1）：

 #include<iostream>//--------1216 HDU   埃拉托色尼筛选法
 #include<stdio.h>
 #include<stdlib.h>
 #include<algorithm>
 #include<math.h>
 #include<string.h>
 #include<vector>
 using namespace std;
 #define Max 40000
 bool b[Max];
 __int64 a[Max], c;
 void fun()
 {
     __int64 i, j, count;
     c = ;
     memset(b, true, sizeof(b));
     b[] = ;
     for (i = ; i < Max; ++i)
     {
         if (b[i])
         {
             a[++c] = i;
             count = i;
             for (j = i+; j < Max; ++j)
             {
                 if (b[j] == true)
                     count--;
                 if (count == )
                 {
                     b[j] = false;
                     count = i;
                 }
             }
         }
     }
 }
 int main()
 {
     fun();
     int n;
     while (scanf("%d", &n) != EOF)
     {
         if (n == )
             break;
         else
             printf("%d\n", a[n]);
     }
     return ;
 }

基本类似的另外一种：

 #include<iostream>
 #include<stdio.h>
 #include<cstdlib>
 #include<string.h>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 #define Max 40000
 __int64 ans[Max];
 __int64 pop[];
 void dfs()
 {
     __int64 i, j, count;
     for (i = ; i < Max; i++)
     {
         ans[i] = i;
     }
     for (i = ; i < Max; i++)
     {
         if (ans[i] != )
         {
             count = i;
             for (j = i + ; j < Max; j++)
             {
                 if (ans[j] != )
                     count--;
                 if (count == )
                 {
                     ans[j] = ;
                     count = i;
                 }
             }
         }

     }
     j = ;
     for (i = ; i <= ; i++)
     {
         while (ans[j] == )
         {
             j += ;
         }
         if (ans[j] != )
         {
             pop[i] = ans[j];
             j += ;
         }
     }
 }
 int main()
 {
     dfs();
     int n;
     while (scanf("%d", &n)&&n)
     {
             printf("%d\n", pop[n]);
     }
     return ;
 }