HDU 1104 Remainder(BFS路径记录+数论）

Posted 2020-10-06 hinata_hajime

Remainder

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4337    Accepted Submission(s): 1095

Problem Description

Coco is a clever boy, who is good at mathematics. However, he is puzzled by a difficult mathematics problem. The problem is: Given three integers N, K and M, N may adds (‘+’) M, subtract (‘-‘) M, multiples (‘*’) M or modulus (‘%’) M (The definition of ‘%’ is given below), and the result will be restored in N. Continue the process above, can you make a situation that “[(the initial value of N) + 1] % K” is equal to “(the current value of N) % K”? If you can, find the minimum steps and what you should do in each step. Please help poor Coco to solve this problem.

You should know that if a = b * q + r (q > 0 and 0 <= r < q), then we have a % q = r.

 

 

Input

There are multiple cases. Each case contains three integers N, K and M (-1000 <= N <= 1000, 1 < K <= 1000, 0 < M <= 1000) in a single line.

The input is terminated with three 0s. This test case is not to be processed.

 

 

Output

For each case, if there is no solution, just print 0. Otherwise, on the first line of the output print the minimum number of steps to make “[(the initial value of N) + 1] % K” is equal to “(the final value of N) % K”. The second line print the operations to do in each step, which consist of ‘+’, ‘-‘, ‘*’ and ‘%’. If there are more than one solution, print the minimum one. (Here we define ‘+’ < ‘-‘ < ‘*’ < ‘%’. And if A = a1a2...ak and B = b1b2...bk are both solutions, we say A < B, if and only if there exists a P such that for i = 1, ..., P-1, ai = bi, and for i = P, ai < bi)

 

 

Sample Input

2 2 2 -1 12 10 0 0 0

 

 

Sample Output

0 2 *+

 

 

Author

Wang Yijie

 

 

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以下有部分引用kuangbin大神

        分析：题目要求最小步数，很容易想到用BFS，这里注意一下

            %与mod的区别：%出来的数有正有负，符号取决于左操作数。。。而mod只能是正（因为a = b * q + r (q > 0 and 0 <= r < q), then we have a mod q = r    中r要大于等于0小于q）。

           所以要用%来计算mod的话就要用这样的公式：a mod b = (a % b + b) % b

            括号里的目的是把左操作数转成正数

 

           要让BFS记录的状态尽量的少，常常会想到在每个地方都%k，记录每次%k后的值

           但是这题会涉及到%k和%m，

           如果记n = 2, m = 8, k =3. 则((n * m)%k % m)%k = 1 而(n * m % m)%k = 0。所以我们不能将每一步记录的值都%k;

          为了不影响后面 所以我们将%km(这里其实用k和m的公倍数就可以了）后的值进行状态记录，关于路径记录则用string进行

         

        

代码如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include <cstring>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
 int  n,k,m;
 int km;
int vis[1000100];
struct node
{
    int x;
    int  step;
    string path;
    int num;
};
queue<node>Q;
node bfs()
{
    int p,q;
  node a,next,fail;
  a.x=n;
  a.step=0;
  Q.push(a);
  q=((n+1)%k+k)%k;
  while(!Q.empty())
  {
   a=Q.front();
   Q.pop();
   p=(a.x%k+k)%k;
    // cout<<p<<" "<<q<<endl;
   if(p==q){
 //   cout<<p<<endl;
   return a;
   }
    next.step=a.step+1;
    next.x=((a.x+m)%km+km)%km;
    next.path=a.path+‘+‘;
    if(!vis[next.x])
    {
        vis[next.x]=1;
        Q.push(next);
    }

     next.x=((a.x-m)%km+km)%km;
    next.path=a.path+‘-‘;
    if(!vis[next.x])
    {
        vis[next.x]=1;
        Q.push(next);
    }

      next.x=(a.x*m%km+km)%km;
    next.path=a.path+‘*‘;
    if(!vis[next.x])
    {
        vis[next.x]=1;
        Q.push(next);
    }
           next.x=(a.x%m+m)%m%km;
    next.path=a.path+‘%‘;
    if(!vis[next.x])
    {
        vis[next.x]=1;
        Q.push(next);
    }

  }
  fail.step=0;
  return fail;
}
int main()
{
  //   freopen("1.out","w",stdout);
    while(scanf("%d%d%d",&n,&k,&m)!=EOF){
    if(n==0&&k==0&&m==0)break;
         km=m*k;
       node ans;
     memset(vis,0,sizeof(vis));
       while(!Q.empty())Q.pop();
       ans=bfs();
       if(ans.step==0)puts("0");
       else{
          printf("%d\n",ans.step);
          cout<<ans.path<<endl;}
   }

   return 0;
}