链接：https://ac.nowcoder.com/acm/contest/296/B 来源：牛客网
Chiaki has a very big sheet of paper. This sheet has a form of rectangle with dimensions 1 x n and numbers from 1 to n was written on each small 1 x 1 grid. Chiaki would like to fold the paper using the following operations:
Fold the sheet of paper at position pi to the right. After this query the leftmost part of the paper with dimensions 1 x pi must be above the rightmost part of the paper with dimensions ).
Fold the sheet of paper at position pi to the left. After this query the rightmost part of the paper with dimensions ) must be above the leftmost part of the paper with dimensions 1 x pi.
After performing the above operations several times, the sheet of paper has dimensions 1 x 1. If we write down the number on each grid from top to bottom, we will have a permutation of n.
Now given a permutation of n, Chiaki would like to know whether it is possible to obtain the permutation using the above operations.
输入描述:
There are multiple test cases. The first line of the input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 106), indicating the length of the paper.
The second line contains n integers a1, a2, …, an, which is a permutation of n, indicating the integers marked in the grids of the resulting sheet of paper from top to bottom.
It’s guaranteed that the sum of n in all test cases will not exceed 106.
输出描述:
For each test case output one line. If it’s possible to obtain the permutation, output ‘‘Yes’’ (without the quotes), otherwise output ‘‘No’’ (without the quotes).
示例1
输入
复制
3
4
2 1 4 3
7
2 5 4 3 6 1 7
4
1 3 2 4
输出
复制
Yes
Yes
No
解法：
考虑两侧的区间，是否全是包含or相离
#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#pragma
#define
#define
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int a[1123456],p[1123456];
int g[1123456];
int main()
{
// freopen("B.in","r",stdin);
// freopen(".out","w",stdout);
int T=read();
while(T--) {
int n=read();
For(i,n) a[i]=read();
For(i,n) p[a[i]]=i;
bool flag=1;
{
memset(g,0,sizeof(int)*(n+1));
For(i,n-1) if(i%2){
int l=p[i],r=p[i+1];if(l>r) swap(l,r);
g[l]=r;
}
stack<int> st;
For(i,n) if(g[i]){
st.push(g[i]);
}else if(!st.empty()) {
if(i==st.top()) st.pop();
}flag&=st.empty();
}
{
memset(g,0,sizeof(int)*(n+1));
For(i,n-1) if(i%2==0){
int l=p[i],r=p[i+1];if(l>r) swap(l,r);
g[l]=r;
}
stack<int> st;
For(i,n) if(g[i]){
st.push(g[i]);
}else if(!st.empty()) {
if(i==st.top()) st.pop();
}flag&=st.empty();
}
puts(flag?"Yes":"No");
}
return 0;
}
