ZOJ 3959 Problem Preparation

It’s time to prepare the problems for the 14th Zhejiang Provincial Collegiate Programming Contest! Almost all members of Zhejiang University programming contest problem setter team brainstorm and code day and night to catch the deadline, and empty bottles of Marjar Cola litter the floor almost everywhere!

To make matters worse, one of the team member fell ill just before the deadline. So you, a brilliant student, are found by the team leader Dai to help the team check the problems’ arrangement.

Now you are given the difficulty score of all problems. Dai introduces you the rules of the arrangement:

1. The number of problems should lie between 10 and 13 (both inclusive).
2. The difficulty scores of the easiest problems (that is to say, the problems with the smallest difficulty scores) should be equal to 1.
3. At least two problems should have their difficulty scores equal to 1.
4. After sorting the problems by their difficulty scores in ascending order, the absolute value of the difference of the difficulty scores between two neighboring problems should be no larger than 2. BUT, if one of the two neighboring problems is the hardest problem, there is no limitation about the difference of the difficulty scores between them. The hardest problem is the problem with the largest difficulty score. It’s guaranteed that there is exactly one hardest problem.

The team members have given you lots of possible arrangements. Please check whether these arrangements obey the rules or not.

给定 N 个数，每个数 ai 代表第 i 题的难度。问能否满足下列条件：

1. 题目数在 10 到 13 题之间，包括 10 题、13 题
2. 最简单题目的难度值应为 1
3. 至少有两个题的题目难度为 1
4. 若将题目难度从小到大排序，相邻题目的难度差不能大于 2 （除了最后一题）。

解题思路

简单模拟，O(N) 判定上述 4 个条件即可。

代码

#include<bits/stdc++.h>using namespace std;const int N = 100 + 10;int n, a[N];bool jug(){    if(n>13 || n < 10 || a[1] != 1 || a[2] != 1)  return false;    for(int i=2;i<n;i++)        if(a[i]-a[i-1] > 2) return false;    return true;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        sort(a+1, a+n+1);        printf("%s\n",jug()?"Yes":"No");    }}