Codeforces 401C Team 贪心法

本题使用贪心法。关键是考贪心策略。同一时候要求要细心，我提交的时候也WA了几次。大意题目就是怎样依照给定的规则排列一个01字符串，引用原题例如以下：

C. Team

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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

有几个坑，代码凝视的地方；推断条件和贪心策略能够依据题意多举例。然后总结出来：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <string>
using std::string;

int main()
{
    int zero, one;
    scanf("%d %d", &zero, &one);
    if (zero > one + 1 ||one > (zero << 1) + 2)///推断无解条件
    {
        printf("-1\n");
    }
    else
    {
        string str;
        while (zero || one)
        {
            if (one + 1 == zero)
            {
                str.insert(str.end(), '0');
                zero--;
            }
            else if ((zero << 1) + 1 <= one)///不要写成(zero<<1)+2 == one
            {
                if (one > 1)/// need to judge first, insure we have enough one
                {
                    str += "11";
                    one -= 2;
                }
                else
                {
                    str += "1";
                    one--;
                }
            }
            else if (str.empty() || (str.back() == '1' && zero > 0))
            {
                str.insert(str.end(), '0');
                zero--;
            }
            else
            {
                str.insert(str.end(), '1');
                one--;
            }
        }
        printf("%s\n", str.c_str());
    }

    return 0;
}