直接枚舉角度 資料比較水吧
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
");
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
",o);
76 return 0;
77 }
78 }
79 if(o==10000)
80 {
81 puts("none");
82 return 0;
83 }
84 }
85 return 0;
86 } View Code