Problem : 1104 ( Remainder ) Judge Status : Accepted
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
using namespace std;
struct cc
{
int n,m;
};
string op[1005];
int step[1005];
int vis[1005],ans,k;
int main()
{
int n,m,ok,flag;
while(scanf("%d%d%d",&n,&k,&m)==3)
{
ok=0;
if(!n&&!m&&!k)break;
memset(step,0,sizeof(step));
memset(vis,0,sizeof(vis));
ans=(n+1+1000*k)%k;
for(int i=0;i<=1001;i++)
op[i].clear();
queue <cc> q;
cc c;
c.m=(n+1000*m)%m;//注意n=-1和n=-1+10(k)做%运算结果不一样，所以事先存n%m;当以后遇到%时c.n=c.m%m；
c.n=(n+1000*k)%k;
q.push(c);
vis[c.n]=1;
while(!q.empty())
{
cc temp=q.front();
q.pop();
for(int i=0;i<4;i++)
{
cc u=temp;
switch(i)
{
case 0:u.n=(u.n+m+1000*k)%k;break;
case 1:u.n=(u.n-m+1000*k)%k;break;
case 2:u.n=(u.n*m+1000*k)%k;u.m=0;break;//u.n*m后，u.n%m==0；以后遇到%运算后u.n==0!=u.m%m;
case 3:u.n=(u.m+1000*k)%k;break;
}
if(!vis[u.n])
{
step[u.n]=step[temp.n]+1;
vis[u.n]=1;
switch(i)
{
case 0:op[u.n]=op[temp.n]+'+';break;
case 1:op[u.n]=op[temp.n]+'-';break;
case 2:op[u.n]=op[temp.n]+'*';break;
case 3:op[u.n]=op[temp.n]+'%';break;
}
q.push(u);
if(u.n==ans){ok=1;flag=u.n;break;}
}
}
if(ok)break;
}
if(!ok)printf("0\n");
else
{
int t=0;
printf("%d\n",step[flag]);
cout<<op[flag]<<endl;
}
}
}