POJ 2034 Anti-prime Sequences

題意：輸入m, n, d。

求出m,m+1,m+2,````m+n的一個排列。使得任意的連續k個數之和都為合數，2<=k<=d。

題解：暴力，依次枚舉第一個數，第二個數····，第n-m+1個數。

#include<cstdio>#include<memory>#include<cstring>using namespace std;const int N = 1005;bool p[N*10], used[N]; // p[0] = false 表示素數int m, n, d, set[N];void prime (){    memset(p,0,sizeof(p));    p[0] = p[1] = 1;    for ( int i = 2; i <= 100; i++ )    {        if ( p[i] ) continue;        for ( int j = 2; i * j < 10001; j++ )            p[i*j] = 1;    }}bool judge ( int index, int value ){    if ( index == 0 )        return true;    int left = index - d + 1;    if ( left < 0 ) left = 0;    int sum = value;    for ( int i = index-1; i >= left; i-- )    {        sum += set[i];        if ( ! p[sum] ) return false;    }    return true;}bool dfs ( int index ){    if ( index == n - m + 1 )        return true;    for ( int i = m; i <= n; i++ )    {        if ( !used[i] && judge(index,i) )        {            set[index] = i;            used[i] = true;            if ( dfs(index+1) ) return true;            used[i] = false;        }    }    return false;}int main(){    prime();    while ( 1 )    {        scanf("%d%d%d",&m,&n,&d);        if ( m + n + d == 0 ) break;        memset(used,0,sizeof(used));        if ( dfs(0) )        {            for ( int i = 0; i < n-m; i++ )                printf("%d,",set[i]);            printf("%d\n",set[n-m]);        }        else printf("No anti-prime sequence exists.\n");    }    return  0;}