CSU1090 数字转换问题[BFS+素数筛选]

1090: Number Transformation

Time Limit: 1 Sec Memory Limit: 128 MB
SUBMIT: 387 Solved: 47
[SUBMIT][STATUS]

Description

In this problem, you are given a pair of integers A and B. You can transform any integer number A to B by adding x to A.This x is an integer number which is a prime below A.Now,your task is to find the minimum number of transformation required to transform S to another integer number T.

Input

Input contains multiple test cases.Each test case contains a pair of integers S and T(0< S < T <= 1000) , one pair of integers per line.

Output

For each pair of input integers S and T you should output the minimum number of transformation needed as Sample output in one line. If it's impossible ,then print 'No path!' without the quotes.

Sample Input

5 7
3 4

Sample Output

Need 1 step(s)
No path!

HINT

Source

CSU Monthly 2012 Jul.

code:

  1 #include <iostream>   
  2 #include <iomanip>   
  3 #include <fstream>   
  4 #include <sstream>   
  5 #include <algorithm>   
  6 #include <string>   
  7 #include <set>   
  8 #include <utility>   
  9 #include <queue>   
 10 #include <stack>   
 11 #include <list>   
 12 #include <vector>   
 13 #include <cstdio>   
 14 #include <cstdlib>   
 15 #include <cstring>   
 16 #include <cmath>   
 17 #include <ctime>   
 18 #include <ctype.h> 
 19 using namespace std;
 20 
 21 int prime[1006];
 22 int s,t;
 23 int vst[1006];
 24 
 25 struct node
 26 {
 27     int num;
 28     int step;
 29     node(){}
 30     node(int a,int b)
 31     {
 32         num=a;
 33         step=b;
 34     }
 35 }Node;
 36 
 37 /*普通筛选法
 38 void isprime()
 39 {
 40     int i,j;
 41     memset(prime,0,sizeof(prime));
 42     for(i=2;i<=1005;i++)
 43     {
 44         if(!prime[i])
 45             for(j=i*i;j<=1005;j+=i)
 46                 prime[j]=1;
 47     }
 48 }*/
 49 
 50 //线性筛选法
 51 bool isPrime[1006];
 52 void isprime()
 53  {  
 54      memset(isPrime,true,sizeof(isPrime));  
 55      memset(prime,0,sizeof(prime));  
 56      int total=0;
 57      for(int i=2;i<=1006;i++)  
 58      {  
 59          if(isPrime[i]) 
 60              prime[total++]=i; 
 61          for(int j=0;j<total&&i*prime[j]<=1006;j++)
 62          {  
 63              isPrime[i*prime[j]]=false; 
 64               if(i%prime[j]==0) 
 65                   break; 
 66          }  
 67      }  
 68  } 
 69  
 70 
 71 int bfs()
 72 {
 73     int i;
 74     int plusnum;    
 75     memset(vst,0,sizeof(vst));
 76     vst[s]=1;
 77     node temp;
 78     temp.num=s;
 79     temp.step=0;
 80     queue<node>Que;
 81     Que.push(temp);
 82     while(!Que.empty())
 83     {
 84         temp=Que.front();
 85         Que.pop();
 86         for(i=2;i<temp.num;i++)
 87         {
 88             if(isPrime[i]==0)
 89                 continue;
 90             plusnum=temp.num+i;
 91             if(plusnum>t)
 92                 break;
 93             if(vst[plusnum])
 94                 continue;
 95             if(plusnum==t)
 96                 return temp.step+1;
 97             vst[plusnum]=1;
 98             Que.push(node(plusnum,temp.step+1));
 99         }
100     }
101     return -1;
102 }
103 
104 int main()
105 {
106     int temp;
107     isprime();
108     while(~scanf("%d%d",&s,&t))
109     {
110         temp=bfs();
111         if(temp==-1)
112             printf("No path!\n");
113         else 
114             printf("Need %d step(s)\n",temp);
115     }
116     return 0;
117 }

普通的筛选法和线性筛选法在这题的时间相差不大，可能是数据量少的原因吧。