HDU - 1016 - Prime Ring Problem(全排列)

题目链接：https://vjudge.net/problem/HDU-1016

题目大意：让你构造一个n个数组成的环，每两个相邻的数之合都是素数

只要用递归算出来所有符合条件的全排列即可

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
#define mst0(a) memset(a, 0, sizeof(a))
#define mstnil(a) memset(a, -1, sizeof(a))
#define mstinf(a) memset(a, 0x3f3f3f3f, sizeof(a))
#define IOS ios::sync_with_stdio(false)
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const double pi = acos(-1.0);
const double eps = 1e-7;
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int NIL = -1;
template<typename T> ll read(T &x){
    x = 0; char ch = getchar(); ll fac = 1;
    while(!isdigit(ch)) { if(ch == '-') fac*=-1; ch = getchar(); }
    while(isdigit(ch)) { x = x*10+ch-48; ch = getchar(); } x*=fac; return x;
}
const int maxn = 1e3+10;
int prime[maxn], kase, nums[maxn], vis[maxn];
bool isprime[maxn] = {1, 1};
void primeMaker() {
    for (int i = 2; i<maxn; ++i) {
        if (!isprime[i])
            prime[kase++] = i;
        for (int j = 0; j<kase && i*prime[j] < maxn; ++j) {
            isprime[prime[j]*i] = true;
            if (!(i%prime[j])) break;
        }
    }
}
void dfs(int pos, int n) {
    if (pos > n && !isprime[nums[1]+nums[n]])
        for (int i = 1; i<=n; ++i)
            printf(i == n ? "%d\n" : "%d ", nums[i]);
    else if (pos > n) return;
    else {
        for (int i = 2; i<=n; ++i)
            if (!vis[i] && !isprime[i+nums[pos-1]]) {
                nums[pos] = i;
                vis[i] = true;
                dfs(pos+1, n);
                vis[i] = false;
            }
    }
}
int main(void) {
    int n, ccase = 1;
    primeMaker();
    while(~scanf("%d", &n)) {
        printf("Case %d:\n", ccase++);
        mst0(vis);
        nums[1] = 1;
        vis[1] = true;
        dfs(2, n);
        putchar(endl);
    }
    return 0;
}