2021.7.27--Benelux Algorithm Programming Contest 2020 补提

I Jigsaw

题目内容：

链接：https://ac.nowcoder.com/acm/contest/18454/I
来源：牛客网

You have found an old jigsaw puzzle in the attic of your house, left behind by the previous occupants. Because you like puzzles, you decide to put this one together. But before you start, you want to know whether this puzzle was left behind for a reason. Maybe it is incomplete? Maybe the box contains pieces from multiple puzzles?
If it looks like a complete puzzle, you also need to know how big your work surface needs to be. Nothing worse than having to start a jigsaw over because you started on a small table.  The box does not tell you the dimensions w × h of the puzzle, but you can quickly count the three types of pieces in the box:
• Corner pieces, which touch two of the edges of the puzzle.
• Edge pieces, which touch one of the edges of the puzzle.
• Center pieces, which touch none of the edges of the puzzle.
Do these pieces add up to a complete jigsaw puzzle? If so, what was the original size of the jigsaw puzzle?
输入描述:

The input consists of:
• One line containing three integers c, e, and m (0 ≤ c,e,m ≤ 10 9 ), the number of corner pieces, edge pieces, and center pieces respectively.

输出描述:

If there exist numbers w and h satisfying w ≥ h ≥ 2 such that the original size of the jigsaw puzzle could have been w×h, then output a single line containing w and h. Otherwise, output “impossible”.
If there are multiple valid solutions, you may output any one of them.

示例1
输入
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4 8 4

输出
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4 4

示例2
输入
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4 10 14

输出
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impossible

示例3
输入
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4 12 6

输出
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impossible

示例4
输入
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4 2048 195063

输出
复制

773 255

题意：有一组拼图，分别给出角块c、边缘块e、中心块m的数量，问能否构成一个完整的拼图。若能则输出该拼图的w和h（w>=h>=2）,不能则输出“impossible”

思路：题目中w和h>=2，所以角块数至少为4，还可以根据拼图经验可得出方程：w*h==c+e+m; (w-2+h-2)*2==e; (w-2)*(h-2)==m;

所以可据此得到代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int c,e,m;
    cin>>c>>e>>m;
    if(c>=4){
        if(e==0&&m==0){
            cout<<2<<" "<<2<<endl;
            return 0;
        }
        else if(e!=0&&m==0&&e%2==0)
        {
            cout<<e/2+2<<" "<<2<<endl;
            return 0;
        }
        else{
            int w,h;
            int t=c+e+m;
            for(w=2;w*w<=c+e+m;w++)
            {
                if(t%w==0)
                {
                    h=t/w;
                    if((w-2+h-2)*2==e&&(w-2)*(h-2)==m)
                    {
                        if(w<h)swap(w,h);
                        cout<<w<<" "<<h<<endl;
                        return 0;
                    }
                }
            }
        }

    }
    cout<<"impossible"<<endl;
}

C Corrupted Contest

题目内容：

链接：https://ac.nowcoder.com/acm/contest/18454/C
来源：牛客网

You are organizing a programming competition in which the rank of a team is fifirst determined by how many problems they have solved. In case of a tie, the team with the lowest time penalty is ranked above the other. However, contrary to the BAPC, the time penalty is equal to t if the latest accepted submission was submitted in the tth minute, or 0 if no problem was solved.
For example, if team A solved their fifirst problem in the 5th minute, their second problem in the 10th minute and their third problem in the 60th minute, then their time penalty is 60. If team B also solved three problems, in the 30th, 40th and 50th minute, their time penalty is 50 and they would rank above team A.
 The contest has fifinished and you would like to enter the fifinal standings. However, due to a corrupted fifile you have lost part of the scoreboard. In particular, the column indicating how many problems each team has solved is gone. You do still have the time penalties of all the teams and know that they are in the right order. You also remember how many problems the contest had. You wonder whether, given this information, it is possible to uniquely reconstruct the number of problems that each team has solved.

输入描述:

The input consists of:
     •One line containing two integers: n (1 ≤ n ≤ 104), the number of teams participating,and p (1 ≤ p ≤ 104), the number of contest problems.
     •n lines with on line i the time score tiin minutes (0 ≤ ti ≤ 106) of the team that is ranked in the ith place.
A positive time score of t indicates that a team has submitted their last accepted submission in the tth minute. A time score of 0 indicates that a team hasn’t solved any problem.
The input always originates from a valid scoreboard.

输出描述:

If it is possible to uniquely reconstruct the scores of all the teams, output n lines containing the number of problems that the ith team has solved on the ith line. Otherwise, output “ambiguous”.

示例1
输入
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9 3
140
75
101
120
30
70
200
0
0

输出
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3
2
2
2
1
1
1
0
0

示例2
输入
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6 3
100
40
40
50
0
0

输出
复制

ambiguous

题意：你正在组织一场编程比赛，其中一个团队的排名首先取决于他们解决了多少问题。

在平局的情况下，罚时最少的球队排在另一队之上。但是，与 BAPC 相反，如果最新接受的提交是在第 t 分钟提交的，则时间惩罚等于 t，如果没有解决问题，则时间惩罚等于 0。

比赛已结束，您想进入决赛排名。但是，由于文件损坏，您丢失了记分板的一部分。特别是，显示每个团队解决了多少问题的列不见了。你仍然有所有球队的时间处罚，并且知道他们的顺序是正确的。你还记得比赛有多少问题。您想知道，根据这些信息，是否可以唯一地重建每个团队已解决的问题数量。

输入描述： 一行包含两个整数：n (1 ≤ n ≤ 104)，参赛队数，p (1 ≤ p ≤ 104)，比赛题数。 • n 行，第i 行是排在第i 位的球队的时间得分tiin 分钟（0 ≤ ti ≤ 106）。 t 的正时间分数表示团队在第 t 分钟提交了他们最后一次接受的提交。时间分数为 0 表示团队没有解决任何问题。 输入始终来自有效的记分板。

输出描述： 如果可以唯一地重建所有团队的分数，则输出 n 行，其中包含第 i 行上第 i 个团队已解决的问题数。否则，输出“ambiguous”。

思路： 可以确定分数时：     1.从榜首到最后一名均为0   2. 不为0时，过题数从m到1或从m到0

不可以确定分数时： 1. 最后一名罚时为0但过题数不为0    2.最后一名罚时不为0，但过题数不为1

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,p;
    cin>>n>>p;
    int s[10010],v[10010],k=0;
    for(int i=0;i<n;i++)
    {
        cin>>s[i];
        if(s[i]!=0)k=1;
    }
    if(k==0)
    {
        while(n--){
            cout<<0<<endl;
        }

    }
    else{
        v[0]=p;
        for(int i=1;i<n;i++)
        {
            if(s[i]<s[i-1])v[i]=--p;
            else v[i]=p;
        }
        if((v[n-1]!=1&&s[n-1]!=0)||(v[n-1]!=0&&s[n-1]==0))//此处不要弄错
            cout<<"ambiguous"<<endl;
        else {
            for(int i=0;i<n;i++)cout<<v[i]<<endl;
        }
    }
}