Poj 2034 Anti-prime Sequences (dfs), pojanti-prime

Poj 2034 Anti-prime Sequences (dfs), pojanti-prime

Http://poj.org/problem? Id = 2034

In this example, [n, m] is given, and the numbers of these intervals are arranged so that the sum of two and three adjacent numbers... d is not a prime number. The smallest Lexicographic Order is output.

Idea: bare dfs. TLE has been used for numerous times because the range of prime numbers in a table is too small. The maximum value should be 10000.

#include <stdio.h>#include <iostream>#include <map>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL long long#define _LL __int64#define eps 1e-8#define PI acos(-1.0)using namespace std;const int maxn = 10010;bool prime[maxn];int vis[1010],ans[1010];int n,m,d;int ok;void init(){    memset(prime,true,sizeof(prime));    prime[0] = prime[1] = false;    for(int i = 2; i <= 10000; i++)    {        if(prime[i] == true)        {            for(int j = i*i; j <= 10000; j += i)                prime[j] = false;        }    }}void dfs(int dep){if(ok == 1)return;if(dep == m-n+2){ok = 1;for(int i = 1; i < dep-1; i++)printf("%d,",ans[i]);printf("%d\n",ans[dep-1]);return;}for(int i = n; i <= m; i++){if(!vis[i]){bool flag = 0;for(int k = 2; k <= d && dep-k>=0; k++){int sum = 0;for(int j = dep-k+1; j <= dep-1; j++)sum += ans[j];if(prime[sum + i] == true)flag = 1;}if(flag == 1)continue;ans[dep] = i;vis[i] = 1;dfs(dep+1);vis[i] = 0;}}}int main(){    init();    while(~scanf("%d %d %d",&n,&m,&d))    {        if(n == 0 && m == 0 && d == 0) break;        memset(vis,0,sizeof(vis));        ok = 0;        dfs(1);        if(ok == 0)printf("No anti-prime sequence exists.\n");    }    return 0;}