您的位置:首页 > 大数据 > 人工智能

2011 Multi-University Training Contest 3 - Host by BIT

2011-08-30 10:59 423 查看



The Lost Traveler

[b]Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 556 Accepted Submission(s): 153

[/b]

Problem Description

Bob is a robot who lives in a two-dimensional world. One day, he decided to go out for a travel. There are two kinds of traveling ways for Bob: to walk forward, and to rotate. When he searched the path with his iPhone 4, he only downloaded part of the data
because of the bad signal. The data just indicated a series of directions for rotation. Bob decided to go on his trip according to this data.

Bob got started at the center of the two-dimensional world (at least he thought he did), and began to move forward. He can walk far enough (but he can't stay at the same place, even in starting point) and then stop to rotate according to the data. The paths
can get intersected.

However, Bob didn't know whether he could get back home or not. So he turned to you for help.

For example, the data indicates like this:

Move forward (He must move for a distance greater than zero. He must not stay and make the next turning without moving forward. The same goes for the following.)

Turn 120 degrees anti-clockwise

Move forward

Turn 120 degrees clockwise

Move forward

Turn 120 degrees clockwise

Move forward

Bob could walk like the figure shows, and he'll get back home.



Input

The first line of the input gives the number of test cases, T. T est cases follow.

Each test case begins with a line containing an integer N,representing the number of degrees. N degrees follow.0<N<100000

Each line contains a number D with a precision of two decimal places, positive for clockwise, negative for anti-clockwise. (-360.00 <= D <= 360.00)

Output

For each case, output one line. If bob can go back to start point, print "Yes". Otherwise, print "No".

Sample Input

2
3
-120.00
120.00
120.00
2
120.00
-120.00


Sample Output

Yes
No


#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#define eps 1e-6
using namespace std;

double a[100010];

int main()
{
int cas;
int n,i;
bool flag;
scanf("%d",&cas);
while(cas--)
{
a[0]=0;
flag=true;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
double b;
scanf("%lf",&b);
a[i]=360.0+a[i-1]+b;
while(a[i]>=360.0)
a[i]-=360.0;
}
sort(a+1,a+n+1);
for(i=1;i<=n;i++)
if(a[i]-a[i-1]>=180)
{
flag=false;
break;
}
if(a
<=180)
flag=false;
if(n==1&&a[1]==180)
flag=true;
if(flag)
printf("Yes\n");
else printf("No\n");
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: 
相关文章推荐