Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
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. — It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
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.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
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 ",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 ",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 "); continue; } sum=bfs(a,b); printf("%d ",sum); } return 0; }