素数环
让你输出满足相邻的相加是素数的序列(注意不要重复)
如:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int stack[25];
int top=-1;
int book[25];
int isprime[40];//初始化为0,不是素数的赋值为1
int n;
int count=0;
void dfs(int step)
{
if(step==n){
int sum=stack[top]+1;//记得算n位置与1位置的和是否为素数
if(isprime[sum]==0)
for(int i=0;i<=top;i++)
printf("%d%c",stack[i],i==top?'\n':' ');
return;
}
for(int i=1;i<=n;i++){
int sum=stack[step-1]+i;
if(book[i]==0&&isprime[sum]==0){
book[i]=1;
stack[++top]=i;
dfs(step+1);
book[i]=0;
top--;
}
}
return;
}
int main()
{
memset(isprime,0,sizeof(isprime));
//存储上素数表
for(int i=2;i<=20;i++){
if(!isprime[i]){
for(int j=i+i;j<=40;j+=i)
isprime[j]=1;
}
}
while(scanf("%d",&n)!=EOF){
memset(book,0,sizeof(book));
top=-1;
printf("Case %d:\n",++count);
//规定第一个数必须为1
book[1]=1;
stack[++top]=1;
dfs(1);
putchar('\n');
}
return 0;
}
后来写的:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;
int n,loc[25],book[25];
int prime[1000];
void dfs(int step){
if(step == n){
if(!prime[loc[n] + loc[1]]){
for(int i=1;i<=n;i++){
printf("%d%c",loc[i],i==n?'\n':' ');
}
}
return;
}
for(int i=1;i<=n;i++){
if(!book[i] && !prime[loc[step] + i]){
loc[step + 1] = i;
book[i] = 1;
dfs(step+1);
book[i] = 0;
}
}
}
int main(){
int k=0;
memset(prime,0,sizeof(prime));
prime[1] = 1;
for(int i = 2;i < sqrt(1000);i++){
if(!prime[i]){
for(int j = i + i;j<1000;j += i){
prime[j] = 1;
}
}
}
while(scanf("%d",&n)!=EOF){
memset(book,0,sizeof(book));
loc[1] = 1;book[1] = 1;
printf("Case %d:\n",++k);
dfs(1);
putchar('\n');
}
return 0;
}