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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#73555#3033. Harry Potter and the Palindromic RadiuszhangbojuAC ✓2121ms40040kbC++141.1kb2023-01-25 20:19:392023-01-25 20:19:40

Judging History

This is the latest submission verdict.

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-01-25 20:19:40]
  • Judged
  • Verdict: AC
  • Time: 2121ms
  • Memory: 40040kb
  • [2023-01-25 20:19:39]
  • Submitted

answer

#include<bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &x)
{
	x=0;short f=1;char c=getchar();
	for(;c<'0'||c>'9';c=getchar()) if(c=='-') f=-1;
	for(;c>='0'&&c<='9';c=getchar()) x=(x<<1)+(x<<3)+(c^48);
	x*=f;return;
}
const int N=2e6+5;
int n;
int h[N];
int a[5][N];
int p[N];
int main()
{
	int T;read(T);
	while(T--)
	{
		read(n);
		for(int i=1;i<=n;++i)
			read(h[i]);
		int idx=0;
		for(int x:{0,1})
			for(int y:{0,1})
			{
				++idx;
				a[idx][1]=x,a[idx][2]=y;
				for(int i=3;i<=n;++i)
					if(h[i-1]) a[idx][i]=a[idx][i-2];
					else a[idx][i]=a[idx][i-2]^1;
				int r=0,c=0;
				for(int i=0;i<=n+2;++i) p[i]=0;
				for(int i=1;i<=n;++i)
				{
					if(r>=i) p[i]=min(r-i,p[2*c-i]);
					else p[i]=0;
					while(i+p[i]+1<=n&&i-p[i]-1>=1)
						if(a[idx][i+p[i]+1]==a[idx][i-p[i]-1]) ++p[i];
						else break;
					if(i+p[i]>r)
						r=i+p[i],c=i;
				}
				bool flag=1;
				for(int i=1;i<=n;++i)
					if(p[i]!=h[i]) flag=0;
				if(!flag) --idx;
			}
		printf("%d\n",idx);
		for(int i=1;i<=idx;++i)
		{
			for(int j=1;j<=n;++j) putchar(a[i][j]+'0');
			putchar('\n');
		}
	}
}

详细

Test #1:

score: 100
Accepted
time: 2121ms
memory: 40040kb

input:

131112
2
0 0
2
0 1
2
0 0
2
1 0
2
0 0
2
0 1
2
0 0
2
0 1
3
0 1 0
3
0 1 1
3
0 0 0
3
0 1 0
3
0 1 0
3
0 2 0
3
0 0 0
3
1 0 0
3
0 0 0
3
1 0 0
3
0 1 0
3
0 2 0
3
0 0 0
3
0 0 1
3
0 1 0
3
0 2 0
4
0 1 1 0
4
0 1 2 0
4
0 0 1 0
4
0 0 1 1
4
0 1 0 0
4
0 1 0 1
4
0 0 0 0
4
0 0 1 0
4
0 0 1 0
4
1 0 1 0
4
0 1 1 0
4
0 1 2...

output:

4
00
01
10
11
0
4
00
01
10
11
0
4
00
01
10
11
0
4
00
01
10
11
0
4
000
010
101
111
0
4
001
011
100
110
4
000
010
101
111
4
000
010
101
111
0
4
001
011
100
110
0
4
001
011
100
110
0
4
000
010
101
111
0
4
001
011
100
110
0
4
000
010
101
111
0
4
0000
0101
1010
1111
0
4
0010
0111
1000
1101
0
4
0001
0100
...

result:

ok 401208 lines