简单搜索 （附素数优化模板F+） （因为是后来补的博客，题意思路直接粘了网上的）

https://vjudge.net/contest/210085#overview

A

在一个给定形状的棋盘（形状可能是不规则的）上面摆放棋子，棋子没有区别。要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列，请编程求解对于给定形状和大小的棋盘，摆放k个棋子的所有可行的摆放方案C。

Input

输入含有多组测试数据。
每组数据的第一行是两个正整数，n k，用一个空格隔开，表示了将在一个n*n的矩阵内描述棋盘，以及摆放棋子的数目。 n <= 8 , k <= n
当为-1 -1时表示输入结束。
随后的n行描述了棋盘的形状：每行有n个字符，其中 # 表示棋盘区域， . 表示空白区域（数据保证不出现多余的空白行或者空白列）。

Output

对于每一组数据，给出一行输出，输出摆放的方案数目C （数据保证C<2^31）。

Sample Input

2 1
#.
.#
4 4
...#
..#.
.#..
#...
-1 -1

Sample Output

2
1


简单dfs，不多解释
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
using namespace std;
int n,k;
    char mp[10][10];
    bool vis[10];
    int flag=0;
void dfs(int res,int ans)
{
    if(ans==k)
    {
        flag++;
        return ;
    }
    if(res>n)return ;
    for(int j=1;j<=n;j++)
    {
        if(mp[res][j]=='#'&&!vis[j])
        {
            vis[j]=true;
            dfs(res+1,ans+1);
            vis[j]=false;
        }

    }
            dfs(res+1,ans);
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF){
    if(n==-1)break;
    getchar();
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)
    {
       scanf("%c",&mp[i][j]);
    }
    getchar();
    }
    memset(vis,false,sizeof(vis));
    flag=0;
    dfs(1,0);
    printf("%d
",flag);
    }
    return 0;
}

B --------------简单题+复杂的数据处理+简单dfs  就直接略了（二维改三维真的头疼）

C

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

 

 

题意： 给数a和b；

要求a-->b     方式：1，+1

                                2，-1

                               3， *2

求最少的步数

 

思路：  bfs等步数跑，，就可以求出最小的步数  （仔细琢磨推算吧，刚开始写的时候脑子还是晕的）

 

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
using namespace std;
int vis[100005]={0};
int o,p;
struct node{
  int x,step;   //x表示值   step表示步数
}s1,s2;    //s1用来存第一次变换，s2用来第二次变换
queue<node>que;
void solve(int x,int step)
{
    s2.x=x;
    vis[x]=1;//标记数值x已经走过了
    s2.step=step+1;
    que.push(s2);
}

int bfs(int a,int b)
{
    int zz;
    s1.x=a;
    s1.step=0;
    que.push(s1);
    while(!que.empty())
    {
       s1=que.front();
       que.pop();
        if(s1.x==b)return s1.step;
         zz=s1.x-1;
         if(zz>=0&&zz<=100005&&!vis[zz])
         {
             solve(zz,s1.step);
         }
         zz=s1.x+1;
         if(zz>=0&&zz<=100005&&!vis[zz])
         {
             solve(zz,s1.step);
         }
         zz=s1.x*2;
         if(zz>=0&&zz<=100005&&!vis[zz])
         {
             solve(zz,s1.step);
         }
    }
    return -1;
}
int main()
{
    while(cin>>o>>p)
    {
        int sun=bfs(o,p);
        cout<<sun<<endl;
    }
    return 0;
}

E

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111


题意：

给一个数n，让你找出一个只有1，0，组成的十进制数，要求是找到的数可以被n整除。

解题思路：

最高位为1， 接下来每一位不是0就是1，双入口bfs

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
using namespace std;
typedef unsigned long long ll;

ll tem,n;
void bfs(int n)
{
    queue<ll>que;
//    while(!que.empty())
//        que.pop();
    que.push(1);
    while(!que.empty())
    {
       tem=que.front();
        que.pop();
        if(tem%n==0){cout<<tem<<endl;break;}
        que.push(tem*10);
        que.push(tem*10+1);
    }
}

int main()
{
    int n;
    while(cin>>n)
    {
        if(n==0)break;
        bfs(n);
    }
    return 0;
}

F

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
8179

The 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

题意：  给定两个素数（四位数），求第一个数经过几次转换可以得到第二个素数。

转换方式：是变换数中某一位的数字（第一位不能为零，其它的变换数字是0~~9），变换之后的数也为素数。

思路：bfs。搜索求最短路径，非常easy就想到广度优先搜索。由于广度优先搜索。第一次搜到得到的步数就是最少的步数。另外打素数表提高推断的时候的效率。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
typedef long long ll;
using namespace std;
struct node{
 int x[4];
 int step;
};
node st,st1,now;
ll ans=0;
int a,b,c,d,sum;
int vis[maxn];
int m,n,t;
bool solve(int n)//素数判定
{
    for(int i=2;i<=sqrt(n);i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}
int bfs()
{
    memset(vis,0,sizeof(vis));
    queue<node>que;
    while(!que.empty())que.pop();
    st.x[0]=m/1000;
    st.x[1]=(m/100)%10;
    st.x[2]=(m/10)%10;
    st.x[3]=m%10;
    st.step=0;
    que.push(st);
    while(!que.empty())
    {
        st1=que.front();
        que.pop();
        if(st1.x[0]==a&&st1.x[1]==b&&st1.x[2]==c&&st1.x[3]==d)
        return st1.step;
        for(int i=0;i<4;i++)
            for(int j=0;j<=9;j++)
        {
            if(i==0&&j==0)continue;  //表示第一位置不能为0
            if(st1.x[i]==j)continue;   //表示如果和原始数值相等的话就不用变换
             now.x[0]=st1.x[0];
            now.x[1]=st1.x[1];
            now.x[2]=st1.x[2];
            now.x[3]=st1.x[3];
            now.step=st1.step;
            now.x[i]=j;
            sum=now.x[0]*1000+now.x[1]*100+now.x[2]*10+now.x[3];
            if(solve(sum)&&!vis[sum])
            {
                vis[sum]=1;
                now.step++;
                que.push(now);
            }
        }
    }
    return -1;
}
int main()
{
    //prime();
    cin>>t;
    while(t--)
    {
        cin>>m>>n;
       a=n/1000;b=(n/100)%10;c=(n/10)%10;d=n%10;
       ll zz=bfs();
       if(zz==-1)
        cout<<"Impossible"<<endl;
       else cout<<zz<<endl;
    }
    return 0;
}





素数筛法的板子（用于判断素数或者求出大区间内所有素数）

//   优化后的筛法就可以避免这种不必要的删去操作 
//#define Max 1000000
//bool prime[Max];
//void IsPrime(){
//     prime[0]=prime[1]=0;prime[2]=1;
//     for(int i=3;i<max;i++)
//        prime[i]=i%2==0?0:1;
//     int t=(int)sqrt(Max*1.0);
//     for(int i=3;i<=t;i++)
//       if(prime[i])
//         for(int j=i*i;j<Max;j+=2*i)//优化 
//            prime[j]=0;
//}


//这就是素数的二次筛法，博士独创~~~~~
//与前两种筛法不同，此种筛法中prime[i]=2*i+3（即：我们只存储奇数，偶数肯定不是素数的） 
//#define Max 1000000
//bool prime[Max>>1];
//void IsPrime(){
//     memset(prime,true,sizeof(prime));
//     int n=Max>>1,m=(int)(sqrt(Max*1.0)/2.0);
//     for(int i=0;i<=m;i++)        
//        if(prime[i])
//          for(int j=2*i*i+6*i+3;j<=n;j+=2*i+3)
//            isprime[j]=false;
//}