Codeforces Round #235 (Div. 2)C. Team

C. Team

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork.

For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that:

• there wouldn't be a pair of any side-adjacent cards with zeroes in a row;
• there wouldn't be a group of three consecutive cards containing numbers one.

Today Vanya brought n cards with zeroes and m cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way.

Input

The first line contains two integers: n (1 ≤ n ≤ 10⁶) — the number of cards containing number 0; m (1 ≤ m ≤ 10⁶) — the number of cards containing number 1.

Output

In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1.

Sample test(s)

input

1 2

output

101

input

4 8

output

110110110101

input

4 10

output

11011011011011

input

1 5

output

-1

题目要求时给定 n 和 m 表示0和1要出现的次数，并且不能连续两个 0 或连续三个 1 出现

用各种if语句写的，写不好

看了大神的代码，看的一知半解，代码如下：

#include <algorithm>#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <map>#define MAXN 1000010#define ll long longusing namespace std;int a[MAXN]; int main(void){int n, m;cin >> n >> m;// L 代表 n 个 0 对应的最少 1 的个数// R 代表 n 个 0 对应的最多 1 的个数 int L = n-1, R = (n+1)*2;//对越界情况的处理 if(m<L || m>R){cout << -1 << endl;return 0;}// i 不从 0 开始是为了后面的语句方便前插 0  for(int i=1; i<n; i++)  a[i] = 1;m -= n-1;//因为 a[] 中已经有 n-1 个 1  for(int i=0; i<=n; i++)while(m>0 && a[i]<2){//m > 0 控制 1 的个数不会超出 m //a[i]<2控制连续的1不会超出2个 m--;a[i]++;}for(int i=0; i<a[0]; ++i)  cout << 1;for(int i=1; i<=n; i++){cout << 0;for(int j=0; j<a[i]; ++j)  cout << 1;}cout << endl;return 0;}