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The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. 1033
1733
3733
3739
3779
8779
8179The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
3
1033 8179
1373 8017
1033 1033
6
7
0
思路：这题的意思也很容易懂，就是给出初始的4位素数后最终的4位素数，通过每次变换个十百千位其中一个（而且变后的数依然是素数）达到最终的4位素数
我认为我的代码比较巧妙的是fun函数。。。。。
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <cmath> using namespace std; #define N 10010 bool prime[N]; int vis[N],step[N]; int aa[]={1,10,100,1000}; void table() { int i,j; memset(prime,true,sizeof(prime)); prime[0]=false; prime[1]=false; for(i=2; i<N; i++) { if(prime[i]) { for(j=2*i; j<N; j+=i) { prime[j]=false; } } } } int fun(int num,int k)//这个函数是让num的K位变成0，例如num=1033，k=1，那么输出的就是1030 { int p[5],q=0,i; while(num>0) { p[q++]=num%10; num/=10; } p[k]=0; for(i=0; i<q; i++) { num+=p[i]*aa[i]; } return num; } int bfs(int a,int b) { int c,temp,i,j,num; memset(vis,0,sizeof(vis)); memset(step,0,sizeof(step)); vis[a]=1; queue<int>q; q.push(a); while(!q.empty()) { c=q.front(); q.pop(); /*printf("%d\n",c);*/ if(c==b) { break; } for(i=0; i<4; i++) { num=fun(c,i); for(j=0; j<=9; j++) { if(i==3&&j==0) { continue; } temp=num+(j*aa[i]); if(prime[temp]&&!vis[temp]) { vis[temp]=1; step[temp]=step[c]+1; /*printf("%d %d\n",temp,step[temp]);*/ q.push(temp); } } } } return step[b]; } int main() { #ifdef CDZSC_OFFLINE freopen("in.txt","r",stdin); #endif int t,a,b,sum; table(); scanf("%d",&t); while(t--) { scanf("%d%d",&a,&b); if(a==b) { printf("0\n"); continue; } sum=bfs(a,b); printf("%d\n",sum); } return 0; }
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原文地址：http://www.cnblogs.com/Wing0624/p/4259625.html
