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
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <queue> 5 #include <string> 6 #include <algorithm> 7 using namespace std; 8 char str[4]={'+','-','*','%'}; 9 int n,k,m; 10 bool vis[200000]; 11 struct node 12 { 13 int num; 14 int step; 15 string st; 16 }st1,st2; 17 18 void bfs() 19 { 20 memset(vis,false,sizeof(vis)); 21 queue<node>q; 22 st1.num=n; 23 st1.step=0; 24 st1.st=""; 25 q.push(st1); 26 vis[((n%k)+k)%k]=true; 27 while(!q.empty()) 28 { 29 st2=q.front(); 30 q.pop(); 31 if(((n+1)%k+k)%k==((st2.num%k)+k)%k) 32 { 33 printf("%d ",st2.step); 34 cout<<st2.st<<endl; 35 return ; 36 } 37 for(int i=0; i<4; i++) 38 { 39 if(i==0) 40 { 41 st1.num=(st2.num+m)%(k*m); 42 st1.st=st2.st+'+'; 43 } 44 else if(i==1) 45 { 46 st1.num=(st2.num-m)%(k*m); 47 st1.st=st2.st+'-'; 48 } 49 else if(i==2) 50 { 51 st1.num=(st2.num*m)%(k*m); 52 st1.st=st2.st+'*'; 53 } 54 else 55 { 56 st1.num=(st2.num%m+m)%m%(k*m); 57 st1.st=st2.st+'%'; 58 } 59 st1.step=st2.step+1; 60 if(!vis[(st1.num%k+k)%k]) 61 { 62 q.push(st1); 63 vis[(st1.num%k+k)%k]=true; 64 } 65 } 66 } 67 printf("0 "); 68 } 69 70 int main() 71 { 72 while(scanf("%d%d%d",&n,&k,&m)!=EOF) 73 { 74 if(n==0&&k==0&&m==0) break; 75 bfs(); 76 } 77 return 0; 78 }