intgcd(int a, int b) { return b ? gcd(b, a % b) : a; }
intmain() { int x, y; cin >> x >> y; constint n = 10000; vector<int> uq(x); uq[0] = n; int sum = n; for (int i = 1; i < x; ++i) { uq[i] = x - i; sum += uq[i]; } int target = n * y; for (int i = 1; i < x; ++i) { int maxuq = uq[i - 1] - 1; int maxadd = maxuq - uq[i]; int taradd = target - sum; if (taradd > maxadd) { uq[i] = maxuq; sum += maxadd; } else { uq[i] += taradd; sum += taradd; break; } } for (int i = 0; i < x - 1; ++i) { int up = uq[i + 1]; int dp = uq[i]; up = dp - up; int gp = gcd(up, dp); up /= gp; dp /= gp; cout << up << " " << dp << "\n"; } cout << "1 1"; return0; }
pointteleport(int x, int y, int k) { int hk = k >> 1; if (k & 1) return {y + hk + 1, x + hk}; return {x + hk, y + hk}; }
voidadd_bfs(int x, int y, int depth, queue<pair<point, int>> &bfs) { if (!bfsed[x][y] && pass[x][y]) { bfsed[x][y] = true; bfs.push({{x, y}, depth}); } }
intmain() { cin >> n >> k; for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) { char sym; cin >> sym; if (sym == '.') pass[i][j] = true; else pass[i][j] = false; } queue<pair<point, int>> bfs; add_bfs(1, 1, 0, bfs); int ans = -1; while (!bfs.empty()) { point np = bfs.front().first; int dep = bfs.front().second; bfs.pop(); int x = np.first, y = np.second; if (x == n && y == n) { ans = dep; break; } int maxk = min(min(2 * (n - x) + 1, 2 * (n - y)), min(n + n - x - y, k)); for (int i = maxk; i > 0; --i) { point tp = teleport(x, y, i); int tx = tp.first, ty = tp.second; if (teled[tx][ty]) break; teled[tx][ty] = true; add_bfs(tx, ty, dep + 1, bfs); } add_bfs(x + 1, y, dep + 1, bfs); add_bfs(x - 1, y, dep + 1, bfs); add_bfs(x, y + 1, dep + 1, bfs); add_bfs(x, y - 1, dep + 1, bfs); } cout << ans; }
using System; // Written by Sheauhaw Jang // 2020-09-22 10:50:57 namespaceCSharp_Week1_A { publicclassCalculate { publicstaticintGet(int a, int b, int type) { switch(type) { case0: return a + b; case1: return a - b; case2: return a * b; case3: return a / b; } return0; } } classProgram { staticvoidMain(string[] args) { int a = Convert.ToInt32(Console.ReadLine()); int b = Convert.ToInt32(Console.ReadLine()); for (int i = 0; i < 4; ++i) { if (i > 0) Console.Write(' '); Console.Write(Calculate.Get(a, b, i)); } } } }
using System; // Written by Sheauhaw Jang // 2020-09-22 11:44:47 namespaceCSharp_Week1_C { publicclassEncode { staticintAddModulo(int x, int p) { return x >= p ? x - p : x; } staticvoid Swap<T>(ref T a, ref T b) { T tmp = b; b = a; a = tmp; } publicstaticint[] DoEncoding(string x) { int[] code = newint[x.Length]; for (int i = 0; i < x.Length; ++i) { code[i] = x[i] - '0'; code[i] = AddModulo(code[i] + 7, 10); } Swap(ref code[0], ref code[2]); Swap(ref code[1], ref code[3]); return code; } publicstaticvoidWriteCode(string x) { int[] code = DoEncoding(x); foreach (int digit in code) Console.Write(digit); } } classProgram { staticvoidMain(string[] args) { string x = Console.ReadLine(); Encode.WriteCode(x); } } }
D. 字符串输入输出
输入一个字符串，按照样例格式输出。
样例：
Alice
Hi Alice,
Welcome to C# 2020!
Best wishes!
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using System; // Written by Sheauhaw Jang // 2020-09-22 11:58:50 namespaceCSharp_Week1_D { classProgram { staticvoidMain(string[] args) { string s = Console.ReadLine(); Console.WriteLine("Hi {0},", s); Console.WriteLine("Welcome to C# 2020!"); Console.WriteLine(); Console.WriteLine("Best wishes!"); } } }
E. 字符大小写转换
输入一个字母，如果它是小写字母则输出它的大写字母，如果它是大写字母输出它的小写字母。
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using System; // Written by Sheauhaw Jang // 2020-09-22 21:28:37 namespaceCSharp_Week1_E { classProgram { staticvoidMain(string[] args) { int s = Console.Read(); if (s >= 'a' && s <= 'z') s = s - 'a' + 'A'; else s = s - 'A' + 'a'; Console.WriteLine(Convert.ToChar(s)); } } }
]]><h1 id="a.-计算两个数的和差商积">A. 计算两个数的和、差、商、积。</h1>
<blockquote>
<p>从键盘分别输入两个不为0的整数（前2行输入），分别输出这两个整数的和 差
积 商。</p>
</blockquote>2020 XJTU 校队夏季训练8 E题题解https://sheauhawjang.github.io/2020/08/31/2020%20Summer%20Training%208E/2020-08-31T15:09:52.000Z2020-09-01T14:27:04.000Z题目来源: 2019-2020 XX Opencup GP of Tokyo: E. Count Modulo 2
#include<bits/stdc++.h> usingnamespacestd; typedeflonglong ll; classllp;// 见我的模板 vector<ll> ps; map<ll, int> cnt; constint mx = 1e5; llp pows[mx], invs[mx]; voiddvd(ll d) { for (ll p = 2; p * p <= d; ++p) if (d % p == 0) { ps.push_back(p); do d /= p; while (d % p == 0); } if (d != 1) ps.push_back(d); } ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } llp solve(ll u, ll v) { if (u == v) return1; if (u > v) swap(u, v); if (v % u != 0) return solve(gcd(u, v), u) * solve(gcd(u, v), v); cnt.clear(); ll up = v / u; ll tcnt = 0; for (ll p : ps) while (up % p == 0) up /= p, ++cnt[p], ++tcnt; llp ans = pows[tcnt]; for (auto x : cnt) ans *= invs[x.second]; return ans; } intmain() { pows[0] = 1; for (int i = 1; i < mx; ++i) pows[i] = pows[i - 1] * i; invs[mx - 1] = ~pows[mx - 1]; for (int i = mx - 2; i >= 0; --i) invs[i] = invs[i + 1] * (i + 1); ll d; scanf("%lld", &d); dvd(d); int q; scanf("%d", &q); for (int t = 0; t < q; ++t) { ll v, u; scanf("%lld%lld", &u, &v); printf("%d\n", (int)solve(v, u)); } return0; }
