F - Prime Path
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.
1033The 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.
1733
3733
3739
3779
8779
8179
Input
3 1033 8179 1373 8017 1033 1033Sample Output
6 7 0
sprintf(s,"%d",n);
参考网址: 点击打开链接
sscanf(s,"%d",&n);
从字符串s中取一个整数值到变量n中。
过程:每一位数字都从1-9枚举一次,如果是素数,就再次枚
举,直到找到目标答案。
#include<stdio.h> #include<stdlib.h> #include<iostream> #include<algorithm> #include<string.h> #include<queue> #include<math.h> using namespace std; const int N=10000; int prime[N]; int s; int e; int vis[N]; int Prime(int x) { for(int i=2;i<=sqrt(x);i++) { if(x%i==0) return 0; } return 1; } int change(int n,int k) // 把n的第k位转化为0; { char s[10]={0}; sprintf(s,"%d",n); //把整数n 打印成一个字符串保存在s 中 //参考网址:http://blog.sina.com.cn/s/blog_980cf62a0100ya0z.html s[k]='0'; sscanf(s,"%d",&n); // 从字符串s中取一个整数值到变量n中。 return n; } int bfs(int s, int e) { int q,w,r; memset(vis,0,sizeof(vis)); queue<int>Q; Q.push(s); vis[s]=1; while(Q.size()) { s=Q.front(); Q.pop(); if(s==e) { return vis[s]-1; } w=1000; for(int i=0;i<4;i++) { q=change(s,i); for(int j=0;j<10;j++) { r=q+j*w; if(prime[r]==1 && vis[r]==0 ) { Q.push(r); vis[r]=vis[s]+1; } } w=w/10; } } return -1; } int main() { int T; for(int i=1000;i<10000;i++) prime[i]=Prime(i); scanf("%d",&T); while(T--) { scanf("%d%d",&s,&e); //不要忘记加 & 符号 细节决定成败 if(bfs(s,e)==-1) { printf("Impossible\n"); } else { printf("%d\n",bfs(s,e)); } } return 0; }