hdu 1104 数论+bfs

Remainder

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2122    Accepted Submission(s): 449

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 *+

 

/*
其他没什么好说的，数字太大需要%（k*m）,这个可以证明就等于对n进行+、-m操作
不影响结果，属于正常操作
*/
#include
#include
#include<string>
using namespace std;
bool vis[1000010];

struct point
{
    int val;
    int step;
    string s;
}p,t;

void bfs(int n,int k,int m)
{
    memset(vis,false,sizeof(vis));
    queue q;
    int s=((n+1)%k+k)%k;
    t.val=n;
    t.step=0;
    t.s="";    
    vis[(n%k+k)%k]=true;
    q.push(t);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(s==(t.val%k+k)%k)
        {
            cout<endl;
            cout<endl;
            return ;
        }
        for(int i=0;i<4;i++)
        {
            p=t;
            p.step++;
            if(i==0)
            {
                p.val=(t.val+m)%(k*m);
                p.s+='+';
            }
            else if(i==1)
            {
                p.val=(t.val-m)%(k*m);
                p.s+='-';
            }
            else if(i==2)
            {
                p.val=(t.val*m)%(k*m);
                p.s+='*';
            }
            else if(i==3)
            {
                p.val=(t.val%m+m)%m%(k*m);
                p.s+='%';
            }
            if(!vis[(p.val%k+k)%k])
            {
                q.push(p);
                vis[(p.val%k+k)%k]=true;
            }
        }
    }
    cout<<0<<endl;
}
int main()
{
    int n,m,k;
    while(cin>>n>>k>>m,k || m || n)
        bfs(n,k,m);
    return 0;
}

 

转载于:https://www.cnblogs.com/xiong-/p/3208275.html