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 "); 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 }