USACO3.23Spinning Wheels

直接枚举角度 数据比较水吧 

 1 /*

 2     ID: shangca2

 3     LANG: C++

 4     TASK: spin

 5  */

 6 #include <iostream>

 7 #include<cstdio>

 8 #include<cstring>

 9 #include<stdlib.h>

10 #include<algorithm>

11 using namespace std;

12 struct node

13 {

14     int l[10],r[10],k,v;

15 }p[10];

16 int f[6][370];

17 int main()

18 {

19     freopen("spin.in","r",stdin);

20     freopen("spin.out","w",stdout);

21     int i,j,o,g;

22     for(i = 1; i <= 5 ; i++)

23     {

24         cin>>p[i].v;

25         cin>>p[i].k;

26         for(j = 1; j <= p[i].k ; j++)

27         {

28             cin>>p[i].l[j]>>p[i].r[j];

29             if(p[i].l[j]+p[i].r[j]>=360)

30             {

31                 for(g = p[i].l[j]; g < 359 ; g++)

32                 f[i][g] = 1;

33                 for(g = 0; g <= (p[i].l[j]+p[i].r[j])%360 ; g++)

34                 f[i][g] = 1;

35             }

36             else

37             for(g = p[i].l[j] ; g <= p[i].l[j]+p[i].r[j] ; g++)

38                 f[i][g] = 1;

39         }

40     }

41     o = 0;

42     for(i = 0 ; i < 360 ; i++)

43     {

44         if(f[1][i]&&f[2][i]&&f[3][i]&&f[4][i]&&f[5][i])

45         {

46             printf("0\n");

47             return 0;

48         }

49     }

50     while(1)

51     {

52         o++;

53         memset(f,0,sizeof(f));

54         for(i = 1; i <= 5 ; i++)

55         {

56             for(j = 1; j <= p[i].k ; j++)

57             {

58                 p[i].l[j] = (p[i].l[j]+p[i].v)%360;

59                 if(p[i].l[j]+p[i].r[j]>=360)

60                 {

61                     for(g = p[i].l[j]; g < 359 ; g++)

62                     f[i][g] = 1;

63                     for(g = 0; g <= (p[i].l[j]+p[i].r[j])%360 ; g++)

64                     f[i][g] = 1;

65                 }

66                 else

67                 for(g = p[i].l[j] ; g <= p[i].l[j]+p[i].r[j] ; g++)

68                 f[i][g] = 1;

69             }

70         }

71         for(i = 0 ; i < 360 ; i++)

72         {

73             if(f[1][i]&&f[2][i]&&f[3][i]&&f[4][i]&&f[5][i])

74             {

75                 printf("%d\n",o);

76                 return 0;

77             }

78         }

79         if(o==10000)

80         {

81             puts("none");

82             return 0;

83         }

84     }

85     return 0;

86 }

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