hdu 1104 Remainder

http://acm.hdu.edu.cn/showproblem.php?pid=1104

a%b=(a%b+b)%b;

题意：开始给了你n, k, m，每次由+m, -m, *m, modm得到新的N，继续对N这样的操作,直到(n+1) mod k== N mod k时结束，并且打印路径

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <string>
#include <algorithm>
using namespace std;
char str[4]={'+','-','*','%'};
int n,k,m;
bool vis[200000];
struct node
{
    int num;
    int step;
    string st;
}st1,st2;

void bfs()
{
    memset(vis,false,sizeof(vis));
    queue<node>q;
    st1.num=n;
    st1.step=0;
    st1.st="";
    q.push(st1);
    vis[((n%k)+k)%k]=true;
    while(!q.empty())
    {
       st2=q.front();
       q.pop();
       if(((n+1)%k+k)%k==((st2.num%k)+k)%k)
       {
           printf("%d
",st2.step);
           cout<<st2.st<<endl;
           return ;
       }
       for(int i=0; i<4; i++)
       {
           if(i==0)
           {
               st1.num=(st2.num+m)%(k*m);
               st1.st=st2.st+'+';
           }
           else if(i==1)
           {
               st1.num=(st2.num-m)%(k*m);
               st1.st=st2.st+'-';
           }
           else if(i==2)
           {
               st1.num=(st2.num*m)%(k*m);
               st1.st=st2.st+'*';
           }
           else
           {
               st1.num=(st2.num%m+m)%m%(k*m);
               st1.st=st2.st+'%';
           }
           st1.step=st2.step+1;
           if(!vis[(st1.num%k+k)%k])
           {
               q.push(st1);
               vis[(st1.num%k+k)%k]=true;
           }
       }
    }
    printf("0
");
}

int main()
{
    while(scanf("%d%d%d",&n,&k,&m)!=EOF)
    {
        if(n==0&&k==0&&m==0) break;
        bfs();
    }
    return 0;
}