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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
#339711#3061. Donut DronemayunfeiRE 7ms4628kbC++142.4kb2024-02-27 20:25:462024-02-27 20:25:47

Judging History

你现在查看的是最新测评结果

  • [2024-02-27 20:25:47]
  • 评测
  • 测评结果:RE
  • 用时:7ms
  • 内存:4628kb
  • [2024-02-27 20:25:46]
  • 提交

answer

#include<bits/stdc++.h>
#define ll long long
#define lll __int128
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
ll rint(ll l,ll r){return uniform_int_distribution<ll>(l,r)(rnd);}
const int maxn=2010;
int n,m,Q,a[maxn][maxn],g[maxn][maxn],nx,ny,f[maxn];
char opt[10];
struct node
{
	int u[maxn];
}t[maxn<<2];
node operator + (node X,node Y)
{
	for(int i=0;i<n;i++) X.u[i]=Y.u[X.u[i]];
	return X;
}
void build(int l,int r,int id)
{
	if(l==r)
	{
		for(int i=0;i<n;i++) t[id].u[i]=g[i][l];
		return;
	}
	int mid=(l+r)>>1;
	build(l,mid,id<<1),build(mid+1,r,(id<<1)|1);
	t[id]=t[id<<1]+t[(id<<1)|1];
}
void modify(int l,int r,int id,int loc)
{
	if(l==r)
	{
		for(int i=0;i<n;i++) t[id].u[i]=g[i][l];
		return;
	}
	int mid=(l+r)>>1;
	if(loc<=mid) modify(l,mid,id<<1,loc);
	else modify(mid+1,r,(id<<1)|1,loc);
	t[id]=t[id<<1]+t[(id<<1)|1];
}
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
			scanf("%d",&a[i][j]);
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			g[i][j]=i;
			if(a[(i-1+n)%n][(j+1)%m]>a[g[i][j]][(j+1)%m]) g[i][j]=(i-1+n)%n;
			if(a[(i+1+n)%n][(j+1)%m]>a[g[i][j]][(j+1)%m]) g[i][j]=(i+1+n)%n;
		}
	}
	build(0,m-1,1);
	scanf("%d",&Q);
	while(Q--)
	{
		scanf("%s",opt);
		if(opt[0]=='m')
		{
			int len;
			scanf("%d",&len);
			while(ny!=0&&len) len--,nx=g[nx][ny],ny=(ny+1)%m;
			if(!len)
			{
				printf("%d %d\n",nx+1,ny+1);
				continue;
			}
			memset(f,0,sizeof(f));
			int cnt=1;
			f[nx]=1;
			while(len>=m)
			{
				nx=t[1].u[nx];
				if(f[nx])
				{
					cnt-=f[nx];
					break;
				}
				f[nx]=++cnt;
				len-=m;
			}
			if(len<m)
			{
				while(len) len--,nx=g[nx][ny],ny=(ny+1)%m;
				printf("%d %d\n",nx+1,ny+1);
				continue;
			}
			len%=(cnt*m);
			while(len>=m) nx=t[1].u[nx],len-=m;
			while(len) len--,nx=g[nx][ny],ny=(ny+1)%m;
			printf("%d %d\n",nx+1,ny+1);
		}
		else
		{
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			x--,y--;
			a[x][y]=z;
			y=(y-1+m)%m;
			for(int i=0;i<n;i++)
			{
				if(a[(i-1+n)%n][(y+1)%m]>a[g[i][y]][(y+1)%m]) g[i][y]=(i-1+n)%n;
				if(a[i][(y+1)%m]>a[g[i][y]][(y+1)%m]) g[i][y]=i;
				if(a[(i+1+n)%n][(y+1)%m]>a[g[i][y]][(y+1)%m]) g[i][y]=(i+1+n)%n;
			}
			modify(0,m-1,1,y);
		}
	}
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3948kb

input:

4 4
1 2 9 3
3 5 4 8
4 3 2 7
5 8 1 6
4
move 1
move 1
change 1 4 100
move 1

output:

4 2
1 3
1 4

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 4036kb

input:

3 4
10 20 30 40
50 60 70 80
90 93 95 99
3
move 4
change 2 1 100
move 4

output:

3 1
2 1

result:

ok 2 lines

Test #3:

score: 0
Accepted
time: 7ms
memory: 4628kb

input:

20 30
8 1 7 4 3 1 2 1 5 5 2 7 6 4 7 1 2 9 1 9 9 9 9 6 4 9 1 6 3 10
7 3 5 6 5 3 5 3 1 9 8 8 7 7 6 7 5 10 9 6 5 3 8 2 7 10 6 8 4 3
6 4 3 1 7 7 6 7 2 7 3 6 8 6 3 2 7 8 10 8 6 5 6 10 2 6 5 10 10 9
2 5 10 9 4 9 3 6 10 8 1 7 3 3 7 3 3 6 7 5 1 1 10 6 8 5 4 2 1 8
5 1 2 8 3 4 2 4 4 9 6 3 1 5 9 1 9 3 8 3 7 10...

output:

19 10
18 19
3 24
3 30
3 2
4 3
3 8
7 14
3 21
2 27
3 7
3 8
6 16
5 17
3 24
2 27
3 29
3 8
7 12
6 16
3 22
3 2
7 12
3 19
1 22
4 25
4 3
3 5
7 12
3 20
1 22
2 24
3 2
6 11
3 20
2 24
4 4
3 5
5 10
6 11
4 18
2 21
3 29
3 2
3 5
4 9
6 16
2 25
1 26
2 1
4 6
7 15
2 24
3 2
7 12
3 19
3 28
5 4
9 11
9 12
11 17
10 19
12 21...

result:

ok 1945 lines

Test #4:

score: -100
Runtime Error

input:

50 50
126812431 909179109 607682292 96000160 425314080 189788877 721251789 103560861 114082307 888028612 277663589 257100764 842807257 327052508 652365304 770138116 384723035 680037089 675501229 509497026 174936063 991259231 761329528 658883078 806406343 741076652 973854314 192609094 398064987 65322...

output:


result: