AtCoder ABC198

本期F为群论题，很有难度。

C - Compass Walking

为了避免精度问题，采用二分推算。但是要小心结果为1的地方。
$R^2?k^2\ge x^2+y^2$

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharm

import 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.stdin
    r, x, y = map(int, fp.readline().split())
    d = x ** 2 + y ** 2
    r = r ** 2
    # r^2 * k^2 >= d^2  (first k)
    lo, hi = 0, 100000001
    while lo < hi:
        mi = (lo + hi) >> 1
        if mi ** 2 * r >= d:
            hi = mi
        else:
            lo = mi + 1
    if hi == 1:
        if r != d:
            hi = 2
    print(hi)


if __name__ == "__main__":
    main()

D - Send More Money

还不错的搜索题，硬做就好了。
一开始思路不好，半小时写了个搜索WA，
因为没有想到答案应该有long long类型。

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <unordered_map>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
const ll inf = ll(1e15);
const ld pi = 3.1415926535897932384626;

string sa, sb, sc;
int na, nb, nc, n;
bool vis[10];
int dig[256];
set<char> ss;
bool ac;
int ansa, ansb, ansc;


void gene() {
    if (dig[sa[sa.size() - 1]] == 0 || dig[sb[sb.size() - 1]] == 0 || dig[sc[sc.size() - 1]] == 0)
    {
        ac = 0;
        return;
    }
    for (int i = 0; i < sa.size(); ++ i) {
        char c = sa[sa.size() - 1 - i];
        ansa = ansa * 10 + dig[c];
    }
    for (int i = 0; i < sb.size(); ++ i) {
        char c = sb[sb.size() - 1 - i];
        ansb = ansb * 10 + dig[c];
    }
    for (int i = 0; i < sc.size(); ++ i) {
        char c = sc[sc.size() - 1 - i];
        ansc = ansc * 10 + dig[c];
    }
}


void dfs(int pos, int row, int c) {
    if (pos == n) {
        if (c == 0) {
            ac = 1;
            gene();
        }
        return;
    }
    if (ac)
        return;
    if (row == 0) {
        int da = pos < na ? dig[sa[pos]] : 0;
        if (da == -1) {
            for (int i = 0; i < 10; ++i) {
                if (!vis[i]) {
                    vis[i] = 1;
                    dig[sa[pos]] = i;
                    dfs(pos, row + 1, c + i);
                    dig[sa[pos]] = -1;
                    vis[i] = 0;
                }
            }
        }
        else {
            dfs(pos, row + 1, c + da);
        }
    }
    else if (row == 1) {
        int db = pos < nb ? dig[sb[pos]] : 0;
        if (db == -1) {
            for (int i = 0; i < 10; ++i) {
                if (!vis[i]) {
                    vis[i] = 1;
                    dig[sb[pos]] = i;
                    dfs(pos, row + 1, c + i);
                    dig[sb[pos]] = -1;
                    vis[i] = 0;
                }
            }
        }
        else {
            dfs(pos, row + 1, c + db);
        }
    }
    else {
        int dc = pos < nc ? dig[sc[pos]] : 0;
        if (dc == -1) {
            int next_c = c / 10;
            dc = c % 10;
            if (!vis[dc]) {
                dig[sc[pos]] = dc;
                vis[dc] = 1;
                dfs(pos + 1, 0, next_c);
                vis[dc] = 0;
                dig[sc[pos]] = -1;
            }
            
        }
        else {
            int tc = c % 10, next_c = c / 10;
            if (tc == dc) {
                dfs(pos + 1, 0, next_c);
            }
        }
    }
}



int main() {
    //freopen("in.txt", "r", stdin);
    cin >> sa >> sb >> sc;
    sa = string(sa.rbegin(), sa.rend());
    sb = string(sb.rbegin(), sb.rend());
    sc = string(sc.rbegin(), sc.rend());
    na = sa.length(), nb = sb.length(), nc = sc.length();
    n = max(max(na, nb), nc);
    memset(dig, 0xff, sizeof(dig));
    for (char x : sa)
        ss.insert(x);
    for (char x : sb)
        ss.insert(x);
    for (char x : sc)
        ss.insert(x);
    if (ss.size() > 10) {
        printf("UNSOLVABLE\n");
        return 0;
    }
    dfs(0, 0, 0);
    if (!ac) {
        printf("UNSOLVABLE\n");
    }
    else {
        printf("%d\n%d\n%d\n", ansa, ansb, ansc);
    }
    return 0;
}

后来看了题解，发现有简便的permutation做法。
p数组是每个字母代表的数字。

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <unordered_map>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
const ll inf = ll(1e15);
const ld pi = 3.1415926535897932384626;

string sa, sb, sc;
set<char> ss;


int main() {
    //freopen("in.txt", "r", stdin);
    cin >> sa >> sb >> sc;
    for (char x : sa)
        ss.insert(x);
    for (char x : sb)
        ss.insert(x);
    for (char x : sc)
        ss.insert(x);
    if (ss.size() > 10) {
        printf("UNSOLVABLE\n");
        return 0;
    }
    vector<char> v(ss.begin(), ss.end());
    int mp[256] = { 0 };
    for (int i = 0; i < v.size(); ++i) {
        mp[v[i]] = i;
    }
    vi p;
    for (int i = 0; i < 10; ++i) p.push_back(i);
    do {
        bool ac = 1;
        ll a = 0, b = 0, c = 0;
        for (int i = 0; i < sa.size(); ++i) {
            a = a * 10 + p[mp[sa[i]]];
            if (i == 0 && p[mp[sa[i]]] == 0) ac = 0;
        }
        for (int i = 0; i < sb.size(); ++i) {
            b = b * 10 + p[mp[sb[i]]];
            if (i == 0 && p[mp[sb[i]]] == 0) ac = 0;
        }
        for (int i = 0; i < sc.size(); ++i) {
            c = c * 10 + p[mp[sc[i]]];
            if (i == 0 && p[mp[sc[i]]] == 0) ac = 0;
        }
        if (ac && a + b == c) {
            printf("%lld\n%lld\n%lld\n", a, b, c);
            return 0;
        }
    } while (next_permutation(p.begin(), p.end()));
    printf("UNSOLVABLE\n");
    return 0;
}

E - Unique Color

搜索
记录一个当前搜索到的颜色数量vis

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <unordered_map>
#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 pair<ll, ll> pll;
typedef vector<int> vi;


int n;
int a[100020];
vi g[100020];
int vis[100020];
int ans = 0;
vi ansl;


void dfs(int u, int fa) {
    if (vis[a[u]] == 0) {
        ansl.push_back(u);
        ans++;
    }
    vis[a[u]] ++;
    for (int v : g[u]) {
        if (v == fa) continue;
        dfs(v, u);
    }
    vis[a[u]] --;
}


int main() {
    //freopen("in.txt", "r", stdin);
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 0);
    //printf("%d\n", ans);
    sort(ansl.begin(), ansl.end());
    for (int x : ansl)
        printf("%d\n", x);
    return 0;
}

F - Cube

Polya定理，暂无