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。
$R^2\ast k^2\ge x^2+y^2$

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