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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#264595 | #5414. Stop, Yesterday Please No More | Frocean | RE | 6ms | 38780kb | C++14 | 2.0kb | 2023-11-25 14:30:41 | 2023-11-25 14:30:41 |
Judging History
answer
#include <iostream>
#include <cstring>
#include <cstdio>
#define N 1500
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 + nowy][m + nowx] = 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 + nowy ; ++ a)
for (int b = 0 ; b < m + nowx ; ++ b) {
int y = min(a + nowy, n + nowy),
x = min(b + nowx, m + nowx);
if (map[y][x] - map[a][x] -
map[y][b] + map[a][b] == jud) ++ ans;
}
printf("%d\n",ans);
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 6ms
memory: 38780kb
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:
0 11 20 99 18 15 34 0 15 0 0 13 14 18 26 16 76 21 108 8 2 2 3 246 22 30 16 21 0 8 10 9 15 5 320 11 7 3 0 0 4 0 11 0 0 14 0 22 36 51 23 4 6 4 2 48 28 8 63 22 49 13 10 4 108 10 18 44 0 15 9 0 4 30 14 99 182 10 14 17 0 66 10 11 28 52 34 56 22 14 56 90 14 0 121 3 48 30 36 13 0 30 7 8 3 11 16 45 20 34 0 ...