2017浙江省赛 B - Problem Preparation ZOJ - 3959
地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3959
題目:
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:
The team members have given you lots of possible arrangements. Please check whether these arrangements obey the rules or not.
Input
There are multiple test cases. The first line of the input is an integer?T?(1 ≤?T?≤ 104), indicating the number of test cases. Then?T?test cases follow.
The first line of each test case contains?one integer?n?(1 ≤?n?≤ 100), indicating the number of problems.
The next line contains?n?integers?s1,?s2, ... ,?sn?(-1000 ≤?si?≤ 1000), indicating the difficulty score of each problem.
We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.
Output
For each test case, output "Yes" (without the quotes) if the arrangement follows the rules, otherwise output "No" (without the quotes).
Sample Input
8 9 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 11 999 1 1 2 3 4 5 6 7 8 9 11 999 1 3 5 7 9 11 13 17 19 21 10 15 1 13 17 1 7 9 5 3 11 13 1 1 1 1 1 1 1 1 1 1 1 1 2 10 2 3 4 5 6 7 8 9 10 11 10 15 1 13 3 6 5 4 7 1 14Sample Output
No No Yes No Yes Yes No NoHint
The first arrangement has 9 problems only, which violates the first rule.
Only one problem in the second and the fourth arrangement has a difficulty score of 1, which violates the third rule.
The easiest problem in the seventh arrangement is a problem with a difficulty score of 2, which violates the second rule.
After sorting the problems of the eighth arrangement by their difficulty scores in ascending order, we can get the sequence 1, 1, 3, 4, 5, 6, 7, 13, 14, 15. We can easily discover that |13 - 7| = 6 > 2. As the problem with a difficulty score of 13 is not the hardest problem (the hardest problem in this arrangement is the problem with a difficulty score of 15), it violates the fourth rule.
思路:
手速題+2,排序后掃一遍就好
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define MP make_pair 6 #define PB push_back 7 typedef long long LL; 8 typedef pair<int,int> PII; 9 const double eps=1e-8; 10 const double pi=acos(-1.0); 11 const int K=1e6+7; 12 const int mod=1e9+7; 13 14 int n,a[K],ans; 15 int main(void) 16 { 17 int t;cin>>t; 18 while(t--) 19 { 20 ans=1; 21 cin>>n; 22 for(int i=0;i<n;i++) 23 scanf("%d",a+i); 24 if(n<10||n>13) 25 { 26 printf("No\n");continue; 27 } 28 sort(a,a+n); 29 if(!(a[0]==1 && a[1]==1)) 30 { 31 printf("No\n");continue; 32 } 33 for(int i=0;i<n-2;i++) 34 if(a[i+1]-a[i]>2) 35 { 36 ans=0;break; 37 } 38 if(ans) 39 printf("Yes\n"); 40 else 41 printf("No\n"); 42 } 43 return 0; 44 }?
轉載于:https://www.cnblogs.com/weeping/p/6764487.html
總結
以上是生活随笔為你收集整理的2017浙江省赛 B - Problem Preparation ZOJ - 3959的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: produces在@requestMap
- 下一篇: [二分]Kayaking Trip