POJ 3126 Prime Path
F - Prime PathPOJ - 3126The 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 of
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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.
Output
Sample Input
Sample Output
— 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. Input
1733
3733
3739
3779
8779
8179
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).
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
3 1033 8179 1373 8017 1033 1033
6 7 0
这道题也是简单搜索目录里的,感觉直接疯狂枚举就能过,结果只能过两个样例,到死都不知道自己错在哪里了,挂一下错误的代码,看看某天会有大佬告知一下吗
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
int main()
{
bool prime[10010];
memset(prime, 1, sizeof(prime));//初始化认为所有数都是素数
prime[0] = prime[1] = 0;//0和1特例标记出来,便于后续判断
prime[2] = 1;
for(int i = 2; i * i < 10010; i ++)
{
if(prime[i])//省去不必要的判断
{
for(int j = i * i; j <= 10010; j += i)//一个数的倍数不是素数
prime[j] = 0;
}
}
int t;
int a, b;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
int sum = 0;
for(int i = a; i <= b; i ++)
{
if(((i/1000==a/1000)&&((i%1000)/100==(a%1000)/100)&&((i%100)/10==(a%100)/10)&&(a%10!=i%10))||
((i/1000==a/1000)&&((i%1000)/100==(a%1000)/100)&&((i%100)/10!=(a%100)/10)&&(a%10==i%10))||
((i/1000==a/1000)&&((i%1000)/100!=(a%1000)/100)&&((i%100)/10==(a%100)/10)&&(a%10==i%10))||
((i/1000!=a/1000)&&((i%1000)/100==(a%1000)/100)&&((i%100)/10==(a%100)/10)&&(a%10==i%10)))//如果i和a有三位数相等
{
if(prime[i])sum++;
}
}
if(sum!=0||a==b)
printf("%d\n",sum);
else printf("Impossible\n");
}
return 0;
}
无脑啊,正常做法应该是bfs,其实和我这个大同小异呀
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
bool prime[10010];
int bfs(int first, int last)
{
bool dis[10010];
queue<int>q;
int v,temp,vtemp;
int count[10010];
int t[4];
memset(dis,0,sizeof(dis));
memset(count,0,sizeof(count));
q.push(first);
dis[first] = 1;
while(!q.empty())
{
v = q.front();
q.pop();
t[0] = v/1000;
t[1] = v%1000/100;
t[2] = v%100/10;
t[3] = v%10;
for(int j = 0; j < 4; j ++)
{
temp = t[j];
for(int i = 0; i < 10; i ++)
if(i!=temp)
{
t[j] = i;
vtemp = t[0]*1000 + t[1]*100 + t[2]*10 + t[3];
if(!dis[vtemp]&&prime[vtemp])
{
count[vtemp] = count[v] + 1;
dis[vtemp] = 1;
q.push(vtemp);
}
if(vtemp == last)return count[vtemp];
}
t[j] = temp;
}
if(v==last)return count[v];
}
return -1;
}
int main()
{
// memset(prime, 1, sizeof(prime));//初始化认为所有数都是素数
// prime[0] = prime[1] = 0;//0和1特例标记出来,便于后续判断
// prime[2] = 1;
// for(int i = 2; i * i < 10010; i ++)
// {
// if(prime[i])//省去不必要的判断
// {
// for(int j = i * i; j <= 10010; j += i)//一个数的倍数不是素数
// prime[j] = 0;
// }
// }
int i, j;
for(i = 1000; i <= 10010; i ++){
for(j = 2; j < i; j ++)
if(i % j == 0)
{
prime[i] = 0;
break;
}
if(i == j)
prime[i] = 1;
}
int t;
int a, b;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
int bl = bfs(a, b);
if(bl==-1)printf("Impossible\n");
else printf("%d\n",bl);
}
return 0;
}
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