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

 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\n",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\n");

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 }

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