USACO Section 3.2 Spinning Wheels - 纯模拟的大水题
主要还是题意~~有点长~感觉确实也讲得不是太清晰... 反正这五个*有那么一个点是相通过去的就ok了...并且要在整数秒时相通..
纯模拟..因为不管速度是多少..反正在360秒后都会转回来...那么就模拟这360秒..每动一秒..判断下有无相通的点...
Program:
/*
ID: zzyzzy12
LANG: C++
TASK: spin
*/
#include<iostream>
#include<istream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<algorithm>
#include<queue>
#define oo 1000000000
#define ll long long
using namespace std;
struct node
{
int speed,num,s[6][3];
}a[6];
int i,j;
void getanswer()
{
int p,k,t,have[380];
for (p=0;p<360;p++)
{
memset(have,0,sizeof(have));
for (i=1;i<=5;i++)
for (j=1;j<=a[i].num;j++)
{
k=(a[i].s[j][0]+p*a[i].speed)%360;
t=(k+a[i].s[j][1])%360;
while (k!=t)
{
have[k]++;
k++;
if (k==360) k=0;
}
have[t]++;
}
for (i=0;i<360;i++)
if (have[i]==5)
{
printf("%d\n",p);
return;
}
}
printf("none\n");
return;
}
int main()
{
freopen("spin.in","r",stdin);
freopen("spin.out","w",stdout);
for (i=1;i<=5;i++)
{
scanf("%d%d",&a[i].speed,&a[i].num);
for (j=1;j<=a[i].num;j++) scanf("%d%d",&a[i].s[j][0],&a[i].s[j][1]);
}
getanswer();
return 0;
}