题目链接:
https://vjudge.net/problem/UVA-524
题意:
给一个n,要求生成1~n的排列,第一个数是1,相邻的两个数的和是素数,包括第一个和最后一个。
题解:
代码:
#include <bits/stdc++.h>
using namespace std;
typedef
long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF =
0x3f3f3f3f;
const ll INFLL =
0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
ll x=
0,f=
1;
char ch=getchar();
while(ch<
'0'||ch>
'9'){
if(ch==
'-')f=-
1;ch=getchar();}
while(ch>=
'0'&&ch<=
'9'){x=x*
10+ch-
'0';ch=getchar();}
return x*f;
}
const int maxn =
50+
10;
int n,isp[maxn],A[maxn],vis[maxn];
bool is_prime(
int x){
int m = sqrt(x);
for(
int i=
2; i<=m; i++)
if(x%i==
0)
return false;
return true;
}
void dfs(
int cur){
if(cur == n+
1 && isp[A[
1]+A[n]]){
printf(
"%d",A[
1]);
for(
int i=
2; i<=n; i++)
printf(
" %d",A[i]);
puts(
"");
return ;
}
for(
int i=
2; i<=n; i++){
if(!vis[i] && (isp[i+A[cur-
1]])){
A[cur] = i;
vis[i] =
1;
dfs(cur+
1);
vis[i] =
0;
}
}
}
int main(){
int cas=
1;
while(scanf(
"%d",&n)!=EOF){
if(cas!=
1) puts(
"");
MS(vis); MS(isp);
for(
int i=
2; i<=n*
2; i++) isp[i] = is_prime(i);
printf(
"Case %d:\n",cas++);
A[
1] =
1;
dfs(
2);
}
return 0;
}
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