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2022-08-10 12:06:00 【51CTO】

题目大概：

输入n组数据,Two four-digit prime numbers for each set of data.Let the first four-digit number change to the second four-digit number,Only one number can be changed per step,The changed number must also be a prime number,Ask at least a few steps to change can be successful.

思路:

用广度优先搜索.

Sift out all prime numbers first,方便判断,Then the four digits are changed one by one,Put the eligible numbers into the queue,until it becomes the second number,输出改变次数.

感想：

数据过多,Make a class of data into a structure,This makes it easier to queue,And if there are multiple queues,内存会不够.

代码：

       
       #include <iostream>
       
       #include <cstdio>
       
       #include <cstring>
       
       #include <queue>
       
       using 
       
       namespace 
       
       std;
       
       int 
       
       sushu[
       
       10001];
       
       int 
       
       biao[
       
       10001];
       
       struct 
       
       q
       
{
       
       int 
       
       bit[
       
       4];
       
       int 
       
       num;
       
       int 
       
       s;
       
};
       
       queue
       
       <
       
       q
       
       >
       
       a;
       
       void 
       
       w()
       
{
       
       int 
       
       i,
       
       j;
       
       for(
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       10001;
       
       i
       
       ++)
       
       sushu[
       
       i]
       
       =
       
       1;
       
       sushu[
       
       0]
       
       =
       
       0;
       
       sushu[
       
       1]
       
       =
       
       0;
       
       for(
       
       i
       
       =
       
       2;
       
       i
       
       <
       
       10001;
       
       i
       
       ++)
       
    {
       
       if(
       
       sushu[
       
       i]
       
       ==
       
       1)
       
        {
       
       for(
       
       j
       
       =
       
       i
       
       +
       
       i;
       
       j
       
       <
       
       10001;
       
       j
       
       +=
       
       i)
       
            {
       
       sushu[
       
       j]
       
       =
       
       0;
       
            }
       
        }
       
    }
       
}
       
       int 
       
       num(
       
       int 
       
       a[],
       
       int 
       
       b,
       
       int 
       
       c)
       
{
       
       int 
       
       s
       
       =
       
       0;
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       4;
       
       i
       
       ++)
       
    {
       
       if(
       
       i
       
       ==
       
       b)
       
       s
       
       =
       
       s
       
       *
       
       10
       
       +
       
       c;
       
       else
       
       s
       
       =
       
       s
       
       *
       
       10
       
       +
       
       a[
       
       i];
       
    }
       
       return 
       
       s;
       
}
       
       void 
       
       bfs(
       
       int 
       
       first,
       
       int 
       
       second)
       
{
       
       int 
       
       i,
       
       j,
       
       k;
       
       memset(
       
       biao,
       
       0,
       
       sizeof(
       
       biao));
       
       q 
       
       f,
       
       r;
       
       f.
       
       num
       
       =
       
       first;
       
       f.
       
       bit[
       
       3]
       
       =
       
       first
       
       %
       
       10;
       
       f.
       
       bit[
       
       2]
       
       =(
       
       first
       
       /
       
       10)
       
       %
       
       10;
       
       f.
       
       bit[
       
       1]
       
       =(
       
       first
       
       /
       
       100)
       
       %
       
       10;
       
       f.
       
       bit[
       
       0]
       
       =(
       
       first
       
       /
       
       1000)
       
       %
       
       10;
       
       f.
       
       s
       
       =
       
       0;
       
       a.
       
       push(
       
       f);
       
       while(
       
       !
       
       a.
       
       empty())
       
    {
       
       r
       
       =
       
       a.
       
       front();
       
       a.
       
       pop();
       
       if(
       
       r.
       
       num
       
       ==
       
       second)
       
        {
       
       cout
       
       <<
       
       r.
       
       s
       
       <<
       
       endl;
       
       break;
       
        }
       
       for(
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       4;
       
       i
       
       ++)
       
{
       
       for(
       
       j
       
       =
       
       0;
       
       j
       
       <=
       
       9;
       
       j
       
       ++)
       
{
       
       if(
       
       j
       
       ==
       
       0
       
       &&
       
       i
       
       ==
       
       0)
       
       continue;
       
       if(
       
       r.
       
       bit[
       
       i]
       
       ==
       
       j)
       
       continue;
       
       else
       
{
       
       int 
       
       i1
       
       =
       
       num(
       
       r.
       
       bit,
       
       i,
       
       j);
       
       if(
       
       sushu[
       
       i1]
       
       &&!
       
       biao[
       
       i1]){
       
       biao[
       
       i1]
       
       =
       
       1;
       
       for(
       
       k
       
       =
       
       0;
       
       k
       
       <
       
       4;
       
       k
       
       ++)
       
{
       
       f.
       
       bit[
       
       k]
       
       =
       
       r.
       
       bit[
       
       k];
       
        }
       
       f.
       
       bit[
       
       i]
       
       =
       
       j;
       
       f.
       
       num
       
       =
       
       i1;
       
       f.
       
       s
       
       =
       
       r.
       
       s
       
       +
       
       1;
       
       a.
       
       push(
       
       f);
       
}
       
}
       
}
       
}
       
}
       
}
       
       int 
       
       main()
       
{
       
       int 
       
       n;
       
       int 
       
       first,
       
       second;
       
       w();
       
       cin
       
       >>
       
       n;
       
       while(
       
       n
       
       --)
       
    {
       
       cin
       
       >>
       
       first
       
       >>
       
       second;
       
       bfs(
       
       first,
       
       second);
       
       while(
       
       !
       
       a.
       
       empty()){
       
       a.
       
       pop();}
       
    }
       
       return 
       
       0;
       
}
      
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https://yzsam.com/2022/222/202208101108135968.html

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