AtCoder abc137

C Green Bin
map计数
D Summer Vacation
想贪心，但没贪成功
应该从后往前考虑，按天计算，维护一个当前可以取到的最大堆

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
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()
    if items[0] == '3.10.6':
        fp = open("in.txt")
    else:
        fp = sys.stdin
    n, k = map(int, fp.readline().split())
    items = []
    for i in range(n):
        a, b = map(int, fp.readline().split())
        items.append([a, b])
    # 从截止日期小到大排序
    # 例如当前的截止日期是1天，那么把1天内可以运行完的都放入堆中
    items.sort(key=lambda x: x[0])
    qu = []
    i = 0
    vis = [0] * n
    ans = 0
    for day in range(1, k + 1):
        while i < n and items[i][0] <= day:
            heapq.heappush(qu, (-items[i][1], i))
            i += 1
        # lazy更新，每一个item进出队列一次
        while len(qu) > 0:
            top = heapq.heappop(qu)
            v, idx = top
            if vis[idx] == 0:
                v = -v
                ans += v
                vis[idx] = 1
                break
    print(ans)


if __name__ == "__main__":
    main()

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E Coins Respawn
很不错的图论水 题
首先易见每条边都减一次cost。
由于有环，只需要用bellman-ford计算最短路
问题是如何判断松弛点对于最后的结果有影响。答案是该松弛点应该在从1到N的路径上。
因此用两个dfs判断每个点是否会被1访问，被N反向访问。

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
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()
    if items[0] == '3.10.6':
        fp = open("in.txt")
    else:
        fp = sys.stdin
    n, m, p = map(int, fp.readline().split())
    g = []
    wg = [[] for _ in range(n)]
    rg = [[] for _ in range(n)]
    dist = [-10 ** 10] * n
    dist[0] = 0
    reach_n = [0] * n
    reach_1 = [0] * n

    def dfs(node):
        reach_n[node] = 1
        for nxt in rg[node]:
            if reach_n[nxt] == 0:
                dfs(nxt)

    def dfs1(node):
        reach_1[node] = 1
        for nxt in wg[node]:
            if reach_1[nxt] == 0:
                dfs1(nxt)

    for i in range(m):
        u, v, w = map(int, fp.readline().split())
        u, v, w = u - 1, v - 1, w - p
        g.append([u, v, w])
        rg[v].append(u)
        wg[u].append(v)

    dfs1(0)
    dfs(n - 1)

    for _ in range(n - 1):
        for i in range(m):
            u, v, w = g[i]
            if dist[v] < dist[u] + w:
                dist[v] = dist[u] + w
    change = 0
    for i in range(m):
        u, v, w = g[i]
        if dist[v] < dist[u] + w and reach_n[v] and reach_1[v]:
            change = 1
    if change:
        print(-1)
    else:
        if dist[n - 1] < 0:
            print(0)
        else:
            print(dist[n - 1])


if __name__ == "__main__":
    main()

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F - Polynomial Construction
费尔马小定理的灵活应用。但是我不会。
已知$a^{p-1}\equiv 1(modp)$
则对于任意$x,p$，
有$1-(x-d{)}^{p-1}\equiv 1x=d$
与$1-(x-d{)}^{p-1}\equiv 0x\ne d$
记上述式子为$f(x,d)$
该式的意义在于，假如我们构造一个多项式
$$F(x)=\sum _{d=1}^{p-1}f(x,d)$$只有当x取到d时，多项式$f(x,d)$才模1，其余情况下多项式$f(x,d)$均为0，也就是整个$F(x)$和为1。
所以我们遍历$a[i]$，当$a[i]=1$时，将$f(x,d)$加入到多项式中。

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
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()
    if items[0] == '3.10.6':
        fp = open("in.txt")
    else:
        fp = sys.stdin
    p = int(fp.readline())
    a = list(map(int, fp.readline().split()))
    b = [0] * p
    cmb = [[0] * (p + 1) for _ in range(p + 1)]
    for j in range(p + 1):
        cmb[0][j] = 1
    for i in range(p + 1):
        cmb[i][0] = 1
        for j in range(1, p + 1):
            cmb[i][j] = (cmb[i - 1][j - 1] + cmb[i - 1][j]) % p
    for i in range(p):
        if a[i] == 1:
            np = 1
            for j in range(p):
                t = -cmb[p - 1][j] * np
                b[p - 1 - j] = (b[p - 1 - j] + t) % p
                np = (np * -i) % p
            b[0] = (b[0] + 1) % p
    print(*b)


if __name__ == "__main__":
    main()

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