hdu1015 + hdu1016 are within ten minutes of deep searching questions

hdu1015: Given a series of values of available sequences, each character is mapped to an integer between 1-26, selected so as to satisfy the ordered number five ab & ² + C ^{. 3} -d ^{. 4} + E ^{. 5} = target.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<map>
 6 using namespace std;
 7 int target;
 8 char s[15];
 9 int vis[28];
10 map<char,int>mp;
11 char ans[10],ch[10];
12 int num[10];
13 int len;
14 bool fin(int *a)
15 {
16     return a[0]-a[1]*a[1]+a[2]*a[2]*a[2]-a[3]*a[3]*a[3]*a[3]+a[4]*a[4]*a[4]*a[4]*a[4]==target;
17 }
18 void init()
19 {
20     memset(vis,0,sizeof(vis));
21     memset(ans,'\0',sizeof(ans));//空串字典序最小 
22     memset(ch,'\0',sizeof(ch));
23     memset(num,0,sizeof(num));
24     char c='A';
25     for(int i=1;i<=26;i++)
26     mp[c++]=i;         //Construction mapping relationship 
27  } 
 28   
29   void DFS ( int CUR) // number of bits currently processed 
30  {
 31 is       IF (CUR == . 5 )
 32       {
 33 is           IF (FIN (NUM) && strcmp (CH, ANS)> 0 )
 34 is           strcpy (ANS, CH);
 35           return ;
 36       }
 37 [           for ( int I = 0 ; I <len; I ++ )
 38 is           {
 39               IF (! VIS [MP [S [I]]])
 40               {
41 is                   VIS [MP [s [i]]] = . 1 ;
 42 is                   CH [CUR] = s [i]; // store character s [i] for comparison with ANS 
43 is                   NUM [CUR] MP = [s [i]] ;   // converted to digital, to determine whether the end 
44 is                   DFS (CUR + . 1 );
 45                   VIS [MP [S [I]]] = 0 ;
 46 is               }
 47          }
 48          return ;
 49  }
 50  int main ()
 51 is  {
 52 is      
53 is      the while (Scanf ( " % S% D " , target &, & S) ==2&&!(target==0&&strcmp(s,"END")==0))
54     {
55         init();
56 //        cout<<"target:"<<target<<"   "<<s<<endl;
57          len=strlen(s);
58         sort(s,s+len);
59         dfs(0);
60         if(!strcmp(ans,""))cout<<"no solution"<<endl;
61         else cout<<ans<<endl;
62     }
63  } 

hdu1016 n number of full-circle configuration arranged such that two adjacent numbers and the first number and a prime number is 1. Preprocessing the first prime number less than 40, then the deep search.

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n;
int a[21];
int isp[40];
int vis[25];
bool isprime(int x)
{
    if(x<2)return false;
    for(int i=2;i*i<=x;i++)
    {
        if(x%i==0)
          return false; 
    } 
    Return  to true ; 
} 
   
void DFS ( int CUR) // Search Depth 
{
     int I;
     IF (CUR == n-&& ISP [(A [cur . 1 ] + A [ 0 ])]) // recursive termination conditions 
    {
         for ( = I 0 ; I <N- . 1 ; I ++ ) 
        the printf ( " % D " , A [I]); 
        the printf ( " % D \ n- " , A [N- . 1 ]); 
    } 
    the else  for (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;       //状态回复 
             }
        }
} 

int main()
{ 
     for(int i=0;i<=40;i++)isp[i]=isprime(i);
    int t=1;
    while(scanf("%d",&n)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        printf("Case %d:\n",t);
        a[0]=1;
        dfs(1);
        printf("\n");
        t++;
    }
 }