Description
Albert, Charles and Mary invented a new version of the classical game Bingo. In traditional Bingo the game is presided over by a non-player known as the caller. At the beginning of the game each player is given a card containing a unique combination of numbers from 0 to N arranged in columns and rows. The caller has a bag containing N + 1 balls, numbered from 0 to N. On each turn, the caller randomly selects a ball from the bag, announces the number of the drawn ball to the players, and sets the ball aside so that it cannot be selected again. Each player searches his card for the called number and marks it if he finds it. The first player who marks a complete pre-announced pattern on the card (for example, a full horizontal line) wins a prize. In the Albert-Charles-Mary version, on each turn, the caller draws a first ball, returns it to the bag, draws a second ball, returns it to the bag, and then calls out the absolute difference between the two ball numbers. To generate even more excitement, before the game started a possibly empty subset of balls is removed from the bag, in such a way that at least two balls remain there. They would like to know if every number from 0 to N may still be called out with the new drawing method considering the balls that were left in the bag.
Input
Each test case is given using exactly two lines. The first line contains two integers N and B. The meaning of N was described above (1 <= N <= 90), while B represents the number of balls which remained in the bag (2 <= B <= N + 1). The second line contains B distinct integers bi, indicating the balls which remained in the bag (0 <= bi <= N). The last test case is followed by a line containing two zeros.
Output
For each test case output a single line containing a single uppercase ‘Y’ if is possible to call out every number from 0 to N, inclusive, or a single uppercase ‘N’ otherwise.
Sample Input
6 7 2 1 3 4 0 6 5 5 4 5 3 0 1 5 3 1 5 0 0 0
Sample Output
Y Y N
HINT
题意：给一段数字a[i]，问每个数和其他数的绝对值的差能不能表示出0-N+1的所有数。
思路：模拟，桶
如果rp不好T了就开小点的数组或者不要memset（v)
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e4;//T了就开小一点
typedef long long LL;
LL v[maxn];
LL k[maxn];
int main(void)
{
LL n,b;
while(cin>>n>>b&&n&&b)
{
memset(v,0,sizeof(v));
memset(k,0,sizeof(k));//这个清空太暴力了在多组下.一般用多少清多少+10
for(int i=1;i<=b;i++)
cin>>v[i];
for(int i=1;i<=b;i++)
for(int j=i+1;j<=b;j++)
{
k[abs(v[i]-v[j])]++;
}
LL flag=1;
for(LL i=1;i<=n;i++)
{
if(!k[i])
{
cout<<"N"<<endl;flag=0;
break;
}
}
if(flag) cout<<"Y"<<endl;
}
return 0;
}