Codeforces 401C Team 贪心法

最新推荐文章于 2022-11-23 09:50:30 发布

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本文探讨了一个利用贪心策略解决特定01字符串排列的问题，通过设定规则确保字符串中不会出现连续两个0或者三个1的情况。文章提供了一段C++代码实现，并详细解释了其中的判断条件及贪心策略。

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本题使用贪心法，关键是考贪心策略，同时要求要细心，我提交的时候也WA了几次，大意题目就是如何按照给定的规则排列一个01字符串，引用原题如下：

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，代码注释的地方；判断条件和贪心策略可以根据题意多举例，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)///判断无解条件
    {
        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;
}