QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#661845#5414. Stop, Yesterday Please No MoreDarkmasterWA 7ms11796kbC++142.6kb2024-10-20 18:28:592024-10-20 18:29:00

Judging History

你现在查看的是最新测评结果

  • [2024-10-20 18:29:00]
  • 评测
  • 测评结果:WA
  • 用时:7ms
  • 内存:11796kb
  • [2024-10-20 18:28:59]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;

const int N = 1e3 + 10, INF = 1e9, base = 1010;
int n, m, k, flag[2 * N][2 * N], sum[2 * N][2 * N];
string s;

void solve(){
    cin >> n >> m >> k;
    for(int i = - n - 1; i <= n; i ++)
        for(int j = - m - 1; j <= m; j ++)
            flag[i + base][j + base] = sum[i + base][j + base] = 0;
    flag[base][base] = 1;
    cin >> s;
    int len = s.size();
    int x = 0, y = 0, minx = 0, miny = 0, maxx = 0, maxy = 0;
    for(int i = 0; i < len; i ++){
        if(s[i] == 'U') x --;
        if(s[i] == 'D') x ++;
        if(s[i] == 'L') y --;
        if(s[i] == 'R') y ++;
        flag[x + base][y + base] = 1;
        minx = min(minx, x);
        miny = min(miny, y);
        maxx = max(maxx, x);
        maxy = max(maxy, y);
    }
    if(minx <= -n || maxx >= n || miny <= -m || maxy >= m){
    	if(k == 0) cout << n * m << '\n';
        else cout << 0 << '\n';
        return;
    }
    int DX1 = 1 + maxx, DX2 = n + minx, DY1 = 1 + maxy, DY2 = m + miny;
//    printf("%d %d %d %d\n", DX1, DX2, DY1, DY2);
    for(int i = - n + base; i <= n + base; i ++)
        for(int j = - m + base; j <= m + base; j ++)
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + flag[i][j];
    int ans = 0;
//    for(int i = -n; i <= n; i ++){
//    	for(int j = -m; j <= m; j ++){
//    		cout << flag[i + base][j + base] << ' ';
//		}
//		cout << '\n';
//	}
//	cout << '\n';
//    for(int i = -n; i <= n; i ++){
//    	for(int j = -m; j <= m; j ++){
//    		cout << sum[i + base][j + base] << ' ';
//		}
//		cout << '\n';
//	}
    for(int i = 1; i <= n; i ++)
    	for(int j = 1; j <= m; j ++){
    		int x = i - DX1 + base, y = j - DY1 + base, xx = i - DX2 + base, yy = j - DY2 + base;
    		if(sum[x][y] - sum[xx - 1][y] - sum[x][yy - 1] + sum[xx - 1][yy - 1] == k){
				ans ++;
//				cout << x - base << ' ' << y - base << ' ' << xx - base << ' ' << yy - base << '\n';
 //				cout << i << ' ' << j << '\n';
			}
		}
//    for(int i = - DX2 + 1 + base; i <= - DX2 + base + n - (DX2 - DX1); i ++)
//        for(int j = - DY2 + 1 + base; j <= - DY2 + base + m - (DY2 - DY1); j ++){
//            int ii = i + (DX2 - DX1), jj = j + (DY2 - DY1);
//            if(sum[ii][jj] - sum[i - 1][jj] - sum[ii][j - 1] + sum[i - 1][j - 1] == k){
//				ans ++;
//				cout << i - base << ' ' << j - base << '\n';
//			}
//        }
    cout << ans << '\n';
}

int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0);
//    cout.tie(0);
    int T;
    cin >> T;
    while(T --) solve();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 7824kb

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
Wrong Answer
time: 7ms
memory: 11796kb

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:

198
11
0
0
0
15
0
157
0
0
0
0
0
18
0
0
0
0
0
0
1
1
3
0
0
0
0
0
0
0
0
9
2
0
320
0
0
3
0
0
0
0
0
0
0
0
0
22
0
51
0
6
6
0
0
48
28
8
0
0
0
0
0
0
0
0
0
44
0
0
0
0
4
30
0
0
105
0
0
17
0
66
0
11
28
0
0
0
0
0
0
90
0
0
0
1
48
0
0
0
0
30
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
48
0
0
99
0
0
0
0
11
0
90
0
0
228
0
30...

result:

wrong answer 1st numbers differ - expected: '228', found: '198'