标签:rip turn solution order get begin ogr cout 数字
#include<iostream>
#include<string.h>
using namespace std;
int judge[41]={0,
0,1,1,0,1,0,1,0,0,0,
1,0,1,0,0,0,1,0,1,0,
0,0,1,0,0,0,0,0,1,0,
1,0,0,0,0,0,1,0,0,0
};//素数的打表
int mark[21];//用作标记是否这个数已经使用过
int y[21];//答案数组
int n;
void dfs(int a){
//如果所有数字确定,那么判断首尾相加是否也是素数,
//若是,则满足条件,输出
if(a==n&&judge[y[1]+y[n]]){
for(int i=1;i<n;i++){
cout<<y[i]<<" ";
}
cout<<y[n]<<endl;
return;
}
//深搜,不断找出未被标记且与此数相加为素数的下一个数
for(int i=1;i<=n;i++){
if(mark[i]==0&&judge[y[a]+i]){
y[a+1]=i;
mark[i]=1;
dfs(a+1);
mark[i]=0;
}
}
}
int main(){
int c=1;
while(cin>>n){
cout<<"Case "<<c++<<":"<<endl;
memset(y,0,sizeof(y));
memset(mark,0,sizeof(mark));
y[1]=1;
mark[1]=1;
dfs(1);
cout<<endl;
}
return 0;
}
HDU-1016 Prime Ring Problem(DFS深搜+打表)
标签:rip turn solution order get begin ogr cout 数字
原文地址:https://www.cnblogs.com/orangecyh/p/9873675.html