Break through in 2 minutes. Just write.
N, X, T = map(int, input().split())
print((N + X - 1) // X * T)
Break through in one and a half minutes. Just write.
N = input()
if sum(int(c) for c in N) % 9 == 0:
print('Yes')
else:
print('No')
It seems that the following was also good in Python. I wonder.
N = int(input())
if N % 9 == 0:
print('Yes')
else:
print('No')
Break through in about 3 minutes. If the previous person is higher, just repeat adding on the table to the same height as the previous person.
N, *A = map(int, open(0).read().split())
result = 0
p = A[0]
for i in range(1, N):
if p >= A[i]:
result += p - A[i]
else:
p = A[i]
print(result)
Break through in 81 minutes, TLE × 2. I was careful that if I implemented it poorly, I would go to the place where I could go with one warp in two times. If there is only one queue, * O * (* HW) in the warp process Since it would be necessary to scan *), if I made two queues and tried to warp only from the cells visited by BFS, the amount of calculation that can be AC was achieved.
package main
import (
"bufio"
"fmt"
"math"
"os"
"strconv"
)
func min(x, y int) int {
if x < y {
return x
}
return y
}
func max(x, y int) int {
if x > y {
return x
}
return y
}
func main() {
defer flush()
H := readInt()
W := readInt()
Ch := readInt() - 1
Cw := readInt() - 1
Dh := readInt() - 1
Dw := readInt() - 1
S := make([]string, H)
for i := 0; i < H; i++ {
S[i] = readString()
}
t := make([][]int, H)
for i := 0; i < H; i++ {
t[i] = make([]int, W)
for j := 0; j < W; j++ {
if S[i][j] == '#' {
t[i][j] = -1
} else {
t[i][j] = math.MaxInt64
}
}
}
warpCount := 0
t[Ch][Cw] = warpCount
q := make([][2]int, 0, 1024)
q = append(q, [2]int{Ch, Cw})
warpq := make([][2]int, 0, 1024)
for len(q) != 0 {
for len(q) != 0 {
warpq = append(warpq, q[0])
h, w := q[0][0], q[0][1]
q = q[1:]
if h-1 >= 0 && t[h-1][w] > warpCount {
q = append(q, [2]int{h - 1, w})
t[h-1][w] = t[h][w]
}
if h+1 < H && t[h+1][w] > warpCount {
q = append(q, [2]int{h + 1, w})
t[h+1][w] = t[h][w]
}
if w-1 >= 0 && t[h][w-1] > warpCount {
q = append(q, [2]int{h, w - 1})
t[h][w-1] = t[h][w]
}
if w+1 < W && t[h][w+1] > warpCount {
q = append(q, [2]int{h, w + 1})
t[h][w+1] = t[h][w]
}
}
if t[Dh][Dw] != math.MaxInt64 {
break
}
warpCount++
for i := 0; i < len(warpq); i++ {
h, w := warpq[i][0], warpq[i][1]
for i := max(0, h-2); i <= min(H-1, h+2); i++ {
for j := max(0, w-2); j <= min(W-1, w+2); j++ {
if t[i][j] <= warpCount {
continue
}
t[i][j] = warpCount
q = append(q, [2]int{i, j})
}
}
}
warpq = warpq[:0]
}
if t[Dh][Dw] == math.MaxInt64 {
println(-1)
} else {
println(t[Dh][Dw])
}
}
const (
ioBufferSize = 1 * 1024 * 1024 // 1 MB
)
var stdinScanner = func() *bufio.Scanner {
result := bufio.NewScanner(os.Stdin)
result.Buffer(make([]byte, ioBufferSize), ioBufferSize)
result.Split(bufio.ScanWords)
return result
}()
func readString() string {
stdinScanner.Scan()
return stdinScanner.Text()
}
func readInt() int {
result, err := strconv.Atoi(readString())
if err != nil {
panic(err)
}
return result
}
var stdoutWriter = bufio.NewWriter(os.Stdout)
func flush() {
stdoutWriter.Flush()
}
func println(args ...interface{}) (int, error) {
return fmt.Fprintln(stdoutWriter, args...)
}
I ended up tuned crisply, but I managed to get it through Python.
from collections import deque
def main():
from sys import stdin
readline = stdin.readline
from builtins import max, min, range
INF = 10 ** 6
H, W = map(int, readline().split())
Ch, Cw = map(lambda x: int(x) - 1, readline().split())
Dh, Dw = map(lambda x: int(x) - 1, readline().split())
S = [readline()[:-1] for _ in range(H)]
t = [[INF] * W for _ in range(H)]
for h in range(H):
th = t[h]
Sh = S[h]
for w in range(W):
if Sh[w] == '#':
th[w] = -1
t[Ch][Cw] = 0
q = deque([(Ch, Cw)])
warp_count = 0
warpq = []
while q:
while q:
warpq.append(q[0])
h, w = q.popleft()
if h - 1 >= 0 and t[h - 1][w] > warp_count:
q.append((h - 1, w))
t[h - 1][w] = warp_count
if h + 1 < H and t[h + 1][w] > warp_count:
q.append((h + 1, w))
t[h + 1][w] = warp_count
if w - 1 >= 0 and t[h][w - 1] > warp_count:
q.append((h, w - 1))
t[h][w - 1] = warp_count
if w + 1 < W and t[h][w + 1] > warp_count:
q.append((h, w + 1))
t[h][w + 1] = warp_count
if t[Dh][Dw] != INF:
break
warp_count += 1
for h, w in warpq:
for i in range(max(0, h - 2), min(H, h + 3)):
ti = t[i]
for j in range(max(0, w - 2), min(W, w + 3)):
if ti[j] > warp_count:
ti[j] = warp_count
q.append((i, j))
warpq.clear()
if t[Dh][Dw] == INF:
print(-1)
else:
print(t[Dh][Dw])
main()
I couldn't break through. If I could solve the D problem 10 minutes earlier ... Or rather, if I skipped D and touched this one ...
Addendum: Obviously this is easier than the D problem. Select the vertical and horizontal with the most blast targets (note that there may be multiple candidates). Double count if there is a blast target at the bomb installation position It is necessary to subtract 1 because it becomes. If you try to remember whether there is a bomb target at the bomb installation position with a simple two-dimensional array, 6 × 10 10 </ sup> bits = 7.5 GB storage capacity is required Therefore, it is memorized by the tuple set of the position.
H, W, M = map(int, input().split())
rows = [0] * H
cols = [0] * W
s = set()
for _ in range(M):
h, w = map(lambda x: int(x) - 1,input().split())
rows[h] += 1
cols[w] += 1
s.add((h, w))
max_rows = max(rows)
max_cols = max(cols)
max_rows_indexes = [i for i in range(H) if rows[i] == max_rows]
max_cols_indexes = [i for i in range(W) if cols[i] == max_cols]
duplicate = True
if M >= len(max_rows_indexes) * len(max_cols_indexes):
for h in max_rows_indexes:
for w in max_cols_indexes:
if (h, w) not in s:
duplicate = False
break
if not duplicate:
break
else:
duplicate = False
if duplicate:
print(max_rows + max_cols - 1)
else:
print(max_rows + max_cols)
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