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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
#264491#5414. Stop, Yesterday Please No MoreFroceanRE 0ms19536kbC++142.0kb2023-11-25 14:11:152023-11-25 14:11:15

Judging History

你现在查看的是最新测评结果

  • [2023-11-25 14:11:15]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:19536kb
  • [2023-11-25 14:11:15]
  • 提交

answer

#include <iostream>
#include <cstring>
#include <cstdio>
#define N 1005
using namespace std;
int map[N << 1][N << 1],t,n,m,k;
string s;
int main() {
	scanf("%d",&t);
	while (t--) {
		scanf("%d%d%d",&n,&m,&k);
		cin >> s;
		memset(map,0,sizeof(map));
		map[n][m] = 1;
		int maxn = 0, minn = 0,
			maxm = 0, minm = 0,
			nowx = 0, nowy = 0;
		for (int a = 0 ; a < s.length() ; ++ a) {
			switch (s[a]) {
				case 'U': --nowy; minn = min(minn, nowy);break;
				case 'D': ++nowy; maxn = max(maxn, nowy);break;
				case 'L': --nowx; minm = min(minm, nowx);break;
				case 'R': ++nowx; maxm = max(maxm, nowx);break;
			}
			if (nowy <= -n || nowy >= n) break;
			if (nowx <= -m || nowx >= m) break;
			map[n + nowx][m + nowy] = 1;
		}
		if (nowy <= -n || nowy >= n || nowx <= -m || nowx >= m) {
			if (k == 0) printf("%d\n",n * m);
			else puts("0"); continue;
		}
		minn = 1 - minn,maxn = n - maxn;
		minm = 1 - minm,maxm = m - maxm;
		nowy = maxn - minn + 1;
		nowx = maxm - minm + 1;
		int ans = 0,jud = nowx * nowy - k;
		if (jud < 0) {puts("0"); continue;}
//		puts("--------------");
//		printf("now minn=%d maxn=%d\n    minm=%d maxm=%d\n",minn,maxn,minm,maxm);
//		for (int x = n + 1,y = m + 1, a = 0 ; a < s.length() ; ++ a) {
//			switch (s[a]) {
//				case 'U': --y;
//				case 'D': ++y;
//				case 'L': --x;
//				case 'R': ++x;
//			}
//			map[x][y] = 1;
//		}
//		puts("---------------");
//		for (int a = 1 ; a <= (n << 1 | 1) ; ++ a,puts(""))
//		for (int b = 1 ; b <= (m << 1 | 1) ; ++ b) printf("%d ",map[a][b]);
		for (int a = 1 ; a <= (n << 1 | 1) ; ++ a)
		for (int b = 1 ; b <= (m << 1 | 1) ; ++ b) map[a][b] += map[a - 1][b];
		for (int a = 1 ; a <= (n << 1 | 1) ; ++ a)
		for (int b = 1 ; b <= (m << 1 | 1) ; ++ b) map[a][b] += map[a][b - 1];
		for (int a = 0 ; a < n ; ++ a)
		for (int b = 0 ; b < m ; ++ b) {
			int y = min(a + nowy, n << 1 | 1), x = min(b + nowx, m << 1 | 1);
			if (map[y][x] - map[a][x] -
				map[y][b] + map[a][b] == jud) ++ ans;
		}
		printf("%d\n",ans);
	}
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 19536kb

input:

3
4 5 3
ULDDRR
4 5 0
UUUUUUU
4 5 10
UUUUUUU

output:

2
20
0

result:

ok 3 number(s): "2 20 0"

Test #2:

score: -100
Runtime Error

input:

1060
19 12 0
UDLDDUUUUDDDLLRDUDUURULUUUDRDUDRDRLRLRLULULLLDLDDRLUUUURUUUDDRLLRUUUDULURUULLRDRLRDDURDUUURRRLURLRUULRRUDURDLUUURDLURDDLUUURDDRLLURRDLRUDLRDRLLRRDRDDLDRURRRLUDULLLRUUDLRRURRDLLRRRDLLRDDDLRLRURURDDDL
11 1 0
UR
3 18 33
UDRLR
17 11 132
RLDRDLDRUU
6 10 13
UULUDDLRDLUUDLDD
1 15 0
D
6 20 50
D...

output:

12
11
3
91
6
15
32
0
13
0
0
4
9
18
8
2
21
3
108
6
1
2
0
50
1
30
12
17
0
0
10
9
14
0
320
4
10
0
0
0
2
0
1
0
0
12
0
22
51
51
13
2
6
2
2
48
28
8
60
1
48
13
5
7
110
1
1
44
0
1
9
0
4
30
14
91
105
5
1

result: