标签：bre logs cti dash cond enc ace one ecif
Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?
Given an integer sequence a1, a2, ..., an of length n. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.
A subsegment is a contiguous slice of the whole sequence. For example, {3, 4, 5} and {1} are subsegments of sequence{1, 2, 3, 4, 5, 6}, while {1, 2, 4} and {7} are not.
The first line of input contains a non-negative integer n (1 ≤ n ≤ 100) — the length of the sequence.
The second line contains n space-separated non-negative integers a1, a2, ..., an (0 ≤ ai ≤ 100) — the elements of the sequence.
Output "Yes" if it‘s possible to fulfill the requirements, and "No" otherwise.
You can output each letter in any case (upper or lower).
3
1 3 5
Yes
5
1 0 1 5 1
Yes
3
4 3 1
No
4
3 9 9 3
No
In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.
In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.
In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.
In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.
题意：把集合分成奇数个，里面的元素也为奇数个，能不能分，当然不破坏顺序
解法：
1 偶数个是没有的
2 奇数个我们只要首尾一个奇数就行
1 #include<bits/stdc++.h> 2 using namespace std; 3 int x[1234]; 4 int main(){ 5 int n; 6 cin>>n; 7 for(int i=1;i<=n;i++){ 8 cin>>x[i]; 9 } 10 if(n%2){ 11 if(x[1]%2&&x[n]%2){ 12 cout<<"Yes"<<endl; 13 }else{ 14 cout<<"No"<<endl; 15 } 16 }else{ 17 cout<<"No"<<endl; 18 } 19 return 0; 20 }
Codeforces Round #431 (Div. 2) A
标签：bre logs cti dash cond enc ace one ecif
原文地址：http://www.cnblogs.com/yinghualuowu/p/7467300.html
