Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches.
The current state of the wall can be respresented by a sequence aa of nn integers, with aiai being the height of the ii -th part of the wall.
Vova can only use 2×12×1 bricks to put in the wall (he has infinite supply of them, however).
Vova can put bricks horizontally on the neighboring parts of the wall of equal height. It means that if for some ii the current height of part ii is the same as for part i+1i+1 , then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 11 of the wall or to the right of part nn of it).
The next paragraph is specific to the version 1 of the problem.
Vova can also put bricks vertically. That means increasing height of any part of the wall by 2.
Vova is a perfectionist, so he considers the wall completed when:
- all parts of the wall has the same height;
- the wall has no empty spaces inside it.
Can Vova complete the wall using any amount of bricks (possibly zero)?
The first line contains a single integer nn (1≤n≤2⋅1051≤n≤2⋅105 ) — the number of parts in the wall.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109 ) — the initial heights of the parts of the wall.
Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero).
Print "NO" otherwise.
5 2 1 1 2 5
YES
3 4 5 3
YES
2 10 10
YES
3 1 2 3
NO
In the first example Vova can put a brick on parts 2 and 3 to make the wall [2,2,2,2,5][2,2,2,2,5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5,5,5,5,5][5,5,5,5,5] .
In the second example Vova can put a brick vertically on part 3 to make the wall [4,5,5][4,5,5] , then horizontally on parts 2 and 3 to make it [4,6,6][4,6,6] and then vertically on part 1 to make it [6,6,6][6,6,6] .
In the third example the wall is already complete.
题意
给出一组数，可以让连续的两个同时加一，也可以让一个数加二，能否用这两种操作使得最后数组的数相同。
分析
用栈一个一个判断就行了，如果栈顶元素和当前数字相同就可以变成比它们大的任意数，如果栈顶元素和当前数组差二的倍数，也可以变成比两个中大的任意数。
/// author:Kissheart /// #include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> #include<vector> #include<stdlib.h> #include<math.h> #include<queue> #include<deque> #include<ctype.h> #include<map> #include<set> #include<stack> #include<string> #define INF 0x3f3f3f3f #define FAST_IO ios::sync_with_stdio(false) const double PI = acos(-1.0); const double eps = 1e-6; const int MAX=2e5+10; const int mod=1e9+7; typedef long long ll; using namespace std; #define gcd(a,b) __gcd(a,b) inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;} inline ll inv1(ll b){return qpow(b,mod-2);} inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;} inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); stack<int>s; int n; int a[MAX]; int main() { while(!s.empty()) s.pop(); scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) { if(s.empty()) { s.push(a[i]); continue; } //printf("%d %d ",s.top(),a[i]); if(s.top()==a[i]) s.pop(); else if(abs(s.top()-a[i])%2==0) s.pop(); else s.push(a[i]); } if(!s.empty()) s.pop(); if(s.empty()) printf("YES "); else printf("NO "); return 0; }