AtCoder ABC186

E题和F题两个绿题。

C - Unlucky 7

单纯的模拟一下。

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)def check(n):t = nwhile t:if t % 10 == 7:return Falset //= 10t = nwhile t:if t % 8 == 7:return Falset //= 8return Truedef main():items = sys.version.split()fp = open("in.txt") if items[0] == "3.10.6" else sys.stdinn = int(fp.readline())ans = 0for i in range(1, n + 1):if check(i):ans += 1print(ans)if __name__ == "__main__":main()

D - Sum of difference

排序

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)def main():items = sys.version.split()fp = open("in.txt") if items[0] == "3.10.6" else sys.stdinn = int(fp.readline())a = list(map(int, fp.readline().split()))a.sort()ans = 0s = 0for i in range(n):ans += a[i] * i - ss += a[i]print(ans)if __name__ == "__main__":main()

E - Throne

如果 $s$ 不能整除 $g c d (k, n)$ ，则无解
$\equiv 0$
$x = (n - s) * in v (k)$
此处的 $in v (k)$ 为 $k$ 在 $n$ 下的逆元
因为 $n$ 未必为质数，因此采用扩展gcd算法计算。

#define _CRT_SECURE_NO_WARNINGS#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;void exgcd(LL a, LL b, LL& x, LL& y) {if (!b) x = 1, y = 0;else exgcd(b, a % b, y, x), y -= a / b * x;
}LL gcd(LL x, LL y) {if (x > y) {LL temp = x;x = y;y = temp;}while (x) {LL temp = x;x = y % x;y = temp;}return y;
}int T;int main() { //freopen("in.txt", "r", stdin);scanf("%d", &T);while (T--) {LL n, s, k;scanf("%lld%lld%lld", &n, &s, &k);LL t0 = gcd(n, k);if (s % t0 != 0) {printf("-1\n");}else {n /= t0;k /= t0;s /= t0;LL k1, y;exgcd(k, n, k1, y);k1 = (k1 % n + n) % n;LL t = (n - s) * k1 % n;printf("%lld\n", t);}}return 0;
}

F - Rook on Grid

用vector记录每列格子的行位置col和每行格子的列位置row，并排序。
遍历每行。
如下面的格子中：

...#
.#..
..#.
#...
....

记r为col[1][0]
第r行之前的策略与第r行之后的策略不同
第r行之前的策略：
对于第i行，找到row[i][0]
之前的格子可以全加。之后的格子需要看上面是否有格子遮挡。row[1][0]后面的格子不需要再计算。
第r行之后的策略：
计算[1…row[1][0]]之间列未被遮挡的数量。
《是否有格子遮挡》可以用BIT维护。

#define _CRT_SECURE_NO_WARNINGS#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;int h, w, m;
vi row[200020];
vi col[200020];
bool vis[200020];
LL c[200020];inline int lowbit(int x) {return x & -x;
}void add(int x, int val) {while (x <= w + 1) {c[x] += val;x += lowbit(x);}
}LL get(int x) {LL ret = 0;while (x > 0) {ret += c[x];x -= lowbit(x);}return ret;
}int main() { //freopen("in.txt", "r", stdin);scanf("%d%d%d", &h, &w, &m);for (int i = 0; i < m; ++i) {int r, c;scanf("%d%d", &r, &c);row[r].push_back(c);col[c].push_back(r);}for (int i = 1; i <= h; ++i){row[i].push_back(w + 1);sort(row[i].begin(), row[i].end());}for (int i = 1; i <= w; ++i) {col[i].push_back(h + 1);sort(col[i].begin(), col[i].end());}LL ans = 0;LL r_limit = col[1][0], c_limit = row[1][0];for (int i = 1; i < r_limit; ++i) {for (int x : row[i]) {if (!vis[x]) {vis[x] = true;add(x, 1);}}LL t = row[i][0] - 1LL;ans += t;if (c_limit > row[i][0]) {t = c_limit - row[i][0] - (get(c_limit) - get(row[i][0]));ans += t;}}for (int i = r_limit; i <= h; ++i) {for (int x : row[i]) {if (!vis[x]) {vis[x] = true;add(x, 1);}}LL t = c_limit - get(c_limit);ans += t;}printf("%lld\n", ans);return 0;
}