sgu137：Funny Strings

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137. Funny Strings

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

Let's consider a string of non-negative integers, containing N elements. Suppose these elements are S₁ S₂ .. S_N, in the order in which they are placed inside the string. Such a string is called'funny' if the string S₁+1 S₂ S₃ .. S_N-1 S_N-1 can be obtained by rotating the first string (to the left or to the right) several times. For instance, the strings 2 2 2 3 and 1 2 1 2 2 are funny, but the string 1 2 1 2 is not. Your task is to find a funny string having N elements, for which the sum of its elements (S₁+S₂+..+S_N) is equal to K.

Input

The input contains two integers: N (2<=N<=1000) and K (1<=K<=30000). Moreover, GCD(N,K)=1 (it can be proven that this is a necessary condition for a string to be funny).

Output

You should output one line containing the elements of the funny string found. These integers should be separated by blanks.

Hint

GCD(A,B) = the greatest common divisor of A and B.
The 'funny' strings are also named Euclid strings in several papers.

Sample Input

9 16

Sample Output

1 2 2 2 1 2 2 2 2

一道被水过的题...（注意“被”字）

首先，rotate指的是S1S2S3...Sn变为SnS1S2S3...Sn-1类似的。

我们分析，若一个串为funny串，为了方便起见，在其后面复制一遍该串，得到新串S'，那么满足序列 S₁+1 S₂ S₃ .. S_N-1 S_N-1 在S'中。

观察可得，S₂ S₃ .. S_{N-1 为不变量，为了方便得到匹配，我们不妨把串的每一项先设为base=K/N，再继续根据需求加上1(总共要加K%N个1)。}

_{怎么加1呢？我们首先来分析K%N=1的情况：}

_{经过初始化后，S={base, base, base, base..., base}，由于}S₁+1与S_N-1均在原串中，不妨将S1=base，Sn=base+1，那么这个串显然满足要求（因为S'中可以从第n项开始找到这个串）。

当K%N=2时，S={base, base, base, base..., base+1}（根据题意和这种构造法必须满足Sn=base+1），还需要加上一个1。

经分析，很容易发现当1加在Sn/2上时，S'中从第n/2项开始可以找到这个串。

当K%N=3时，依此类推...

由此我们得出加1的规律，设mod=K%N，那么S（1+i*N/mod）=base+1，1+i*N/mod∈(1,n)，且Sn=base+1，其余的Sk=base。

#include <cstdio>using namespace std;int N, K;int A[1005] = {0};int main(){  int base = 0, mod = 0;  scanf("%d%d", &N, &K);    base = K/N;  mod = K%N;  for(int i = 1; i <= N; ++i) A[i] = base;  for(int i = 1; i*N/mod <= N; ++i) A[1+i*N/mod]++;  A[N]++;    printf("%d", A[1]);  for(int i = 2; i <= N; ++i) printf(" %d", A[i]);  printf("\n");  return 0;}

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