ACcode

A. Asia區(qū)域賽

讀題目輸出獎(jiǎng)牌之和就行，我把冠亞軍也記成獎(jiǎng)牌了。。

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 1001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std; const int mod = 1e9 + 7;int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int ans = 29 + 87 + 58 + 11 + 1 + 1;cout << ans << endl;return 0; }

B. Asia區(qū)域制

二進(jìn)制轉(zhuǎn)化為十進(jìn)制，當(dāng)成字符串搞搞 “4位一并”

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 1001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std; const int mod = 1e9 + 7;int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int T;cin >> T;map<int, char> mp;mp[10] = 'a';mp[11] = 'b';mp[12] = 'c';mp[13] = 'd';mp[14] = 'e';mp[15] = 'f';while (T--) {string s;cin >> s;string t;int len = s.size();int last = len % 4;for (int i = 0; last && i < 4 - last; ++i) {s = "0" + s;}len = s.size();int tmp = 0;int cnt = 0;for (int i = 0; i < len; ++i) {++tmp;cnt = cnt * 2 + s[i] - '0';if (tmp % 4 == 0) {// cout << tmp << endl;if (cnt <= 9) t += '0' + cnt;else t += mp[cnt];cnt = 0;}}cout << t << endl;}return 0; }

C. Asia區(qū)域?qū)m

每行每列只有一個(gè)障礙，只有存在一條對(duì)角線才可能把起點(diǎn)封住，讀入點(diǎn)的時(shí)候預(yù)處理每種對(duì)角線的點(diǎn)數(shù)，如果存在點(diǎn)數(shù)合理就是NO，否者就是曼哈頓距離

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 10001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std;int n, m; int sum[maxn];int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int T;cin >> T;while (T--) {mem(sum, 0);cin >> n >> m;for (int i = 0, x, y; i < m; ++i) {cin >> x >> y;sum[x+y-1]++;}int flag = 1;for (int i = 1; i <= n; ++i) {if (sum[i] == i) flag = 0;}if (flag) cout << "Yes " << 2*n - 2 << endl;else cout << "No\n";}return 0; }

D. Asia區(qū)域陣

當(dāng)時(shí)讀題的時(shí)候感覺好難，數(shù)據(jù)給的很小，但是暴力的話感覺很難寫，于是就放棄了。
補(bǔ)個(gè)暴力，枚舉矩陣的起點(diǎn)然后計(jì)算以該點(diǎn)能擴(kuò)展的最大矩陣，列擴(kuò)展的時(shí)候需要標(biāo)記，因?yàn)橹蟮牧幸欢ㄐ∮诘扔诋?dāng)前所能擴(kuò)展的最小列

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 1000001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std;char a[101][101]; int R[101][91], C[101][91]; int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int T;scanf("%d", &T);while (T--) {int n, m;scanf("%d %d", &n, &m);for (int i = 0; i < n; ++i) scanf("%s", a[i]);int ans = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {mem(R, 0);mem(C, 0);int c = m;for (int p = i; p < n; ++p) {for (int k = j; k < c; ++k) {int t = a[p][k];if (R[p][t] || C[k][t]) {c = k;break;}ans = max(ans, (p-i+1)*(k-j+1));R[p][t] = 1;C[k][t] = 1;}}}}printf("%d\n", ans);}return 0; }

E. Mo的游戲

統(tǒng)計(jì)每個(gè)字符出現(xiàn)的位置，然后for一遍，找最小的差（最小的差一定相鄰）

#include <bits/stdc++.h> using namespace std; typedef long long LL;vector<int> g[1000];int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); string s;cin >> s;int len = s.size();for (int i = 0; i < len; ++i) {int t = s[i];g[t].push_back(i);}for (int i = 'a'; i <= 'z'; ++i) {if (g[i].empty()) continue;int ll = g[i].size();if (ll == 1) {cout << char(i) << ":0" << endl;}else {int tmp = g[i][1] - g[i][0];for (int j = 2; j < ll; ++j) {tmp = min(tmp, g[i][j] - g[i][j-1]);}cout << char(i) << ":" << len - tmp << endl;}}for (int i = 'A'; i <= 'Z'; ++i) {if (g[i].empty()) continue;int ll = g[i].size();if (ll == 1) {cout << char(i) << ":0" << endl;}else {int tmp = g[i][1] - g[i][0];for (int j = 2; j < ll; ++j) {tmp = min(tmp, g[i][j] - g[i][j-1]);}cout << char(i) << ":" << len - tmp << endl;}}return 0; }

F. Mo的極限

字符串處理統(tǒng)計(jì)次方和對(duì)應(yīng)的系數(shù)，最后找最大的判斷

• 分子上的系數(shù)可能全部為0
• 可能存在相同指數(shù)的多項(xiàng)式，需要去重
• 最簡(jiǎn)分?jǐn)?shù)的分母是負(fù)數(shù)
• 最簡(jiǎn)分?jǐn)?shù)的分母是1

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 1001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std; const int mod = 1e9 + 7; struct ac{int d, x;bool operator < (const ac &t) {return x > t.x;} }a[maxn], b[maxn];void solve(string s, int &d, int &x) {int flag = 1;int len = s.size();if (s[0] == '-') flag = -1;int now = 0;if (s[0] == '+' || s[0] == '-') now = 1;int xx;for (int i = 0; i < len; ++i) {if (s[i] == '^') {xx = i + 1;break;}}int num = 0;for (int i = now; s[i] != 'x'; ++i) {num = num * 10 + s[i] - '0';}d = num * flag;num = 0;for (int i = xx; i < len; ++i) {num = num * 10 + s[i] - '0';}x = num; }int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); string s, t, tmp;cin >> s >> t;int lens, lent;int ls = s.size();int lt = t.size();int now = 1;map<int,int> mp;//處理第一個(gè)多項(xiàng)式for (int i = 0; i < ls; ++i) {if (i == ls-1 || (i && (s[i] == '+' || s[i] == '-'))) {if (i == ls-1) tmp += s[i];int d, x;solve(tmp, d, x);if (mp[x]) {//去重a[mp[x]].d += d;tmp = s[i];continue;}mp[x] = now;a[now].d = d;a[now++].x = x;tmp = s[i];}else tmp += s[i];}mp.clear();lens = now;now = 1;tmp = "";// 處理第二個(gè)多項(xiàng)式for (int i = 0; i < lt; ++i) {if (i == lt-1 || (i && (t[i] == '+' || t[i] == '-'))) {if (i == lt-1) tmp += t[i];int d, x;solve(tmp, d, x);if (mp[x]) {b[mp[x]].d += d;tmp = t[i];continue;}mp[x]=now;b[now].d = d;b[now++].x = x;tmp = t[i];}else tmp += t[i];}lent = now;sort(a+1, a + lens+1);sort(b+1, b + lent+1);int n, m = -1, a0, b0;for (int i = 1; i < lens; ++i) {if (a[i].d != 0) {a0 = a[i].d; m = a[i].x;break;}}for (int i = 1; i < lent; ++i) {if (b[i].d != 0) {b0 = b[i].d; n = b[i].x;break;}} if (n > m) cout << 0 << endl;else if (n < m) cout << "oo\n";else {int tt = __gcd(abs(a0), abs(b0));a0 /= tt;b0 /= tt;if (b0 < 0) {b0 *= -1;a0 *= -1;}if (b0 == 1) cout << a0 << endl;else cout << a0 << "/" << b0 << endl;}return 0; }

G. Mo的數(shù)學(xué)

容斥，初始化$ans=n!$，然后篩m的素因子，$1?n$中與$m$互質(zhì)的數(shù)一定不含有$m$的素因子，容器奇數(shù)個(gè)素因子除掉，偶數(shù)個(gè)素因子乘上，每個(gè)素因子最多有$\frac{n}{t}$個(gè)
$$\frac{ans}{1t\times 2t\times ...\times \frac{n}{t}t}=\frac{ans}{\frac{n}{t}!t^{\frac{n}{t}}}$$
$$ans\times 1t\times 2t\times ...\times \frac{n}{t}t=ans\times \frac{n}{t}!t^{\frac{n}{t}}$$

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 1000001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std; LL mod = 1e9 + 7; LL jie[maxn]; LL exgcd(LL a, LL b, LL &x, LL &y) {LL d = a;if (b == 0) x = 1,y = 0;else {d = exgcd(b, a%b, y, x);y -= a / b * x;}return d; }LL inv(LL a, LL mod) {LL x, y;LL d = exgcd(a, mod, x, y);return (d == 1) ? ((x + mod) % mod) : -1; }LL pow(LL x, LL a) {LL ans = 1;while (a) {if (a&1) ans = ans * x % mod;x = x * x % mod;a >>= 1;}return ans; }int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int n, m;jie[1] = jie[0] = 1;for (LL i = 2; i < maxn; ++i) {jie[i] = jie[i-1] * i % mod;}while (cin >> n >> m) {vector<LL> su;for (LL i = 2; i * i <= m; ++i) {if (m % i == 0) {su.push_back(i);while (m % i == 0) m /= i;}}if (m > 1) su.push_back(m);LL ans = jie[n];int len = su.size();for (int i = 1; i < (1<<len); ++i) {int cnt = 0;LL t = 1; for (int j = 0; (1<<j) <= i; ++j) {if (i & (1<<j)) {cnt++;t = t * su[j] % mod;}}LL x = jie[n/t] * pow(t, n/t) % mod;if (cnt & 1) ans = ans * inv(x, mod) % mod;else ans = ans * x % mod;}cout << ans << endl; }return 0; }

H. Mo的面積

只有兩個(gè)矩形的矩形面積并，分類討論相交矩形的長(zhǎng)和寬

#include <bits/stdc++.h> using namespace std; typedef long long LL; struct ac{int a, b, c, d; };int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); ac x, y;cin >> x.a >> x.b >> x.c >> x.d;cin >> y.a >> y.b >> y.c >> y.d;if (x.a > y.a) swap(x, y);int ans = 0;ans = (x.c - x.a) * (x.d - x.b);ans += (y.c - y.a) * (y.d - y.b);int l;if (y.a >= x.a && y.c <= x.c) l = y.c - y.a;if (y.a > x.a && y.c > x.c && y.a < x.c) l = x.c - y.a;if (y.a < x.a && y.c > x.a && y.c < x.c) l = y.c - x.a;if (y.a <= x.a && y.c >= x.c) l = x.c - x.a;if (y.c <= x.a || y.a >= x.c) l = 0;int r;if (y.b >= x.d || y.d <= x.b) r = 0;if (y.d >= x.d && y.b <= x.b) r = x.d - x.b;if (y.d <= x.d && y.b >= x.b) r = y.d - y.b;if (y.d > x.b && y.b < x.b && y.d < x.d) r = y.d - x.b;if (y.b > x.b && y.d > x.d && y.b < x.d) r = x.d - y.b;ans -= l * r;cout << ans << endl;return 0; }

J. 簡(jiǎn)單遞歸

$f[n]=2f[n?1]+\frac{f[n?2]}{2}$

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 105; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std; int mod = 1e9 + 7; LL f[1000006]; int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int T;cin >> T;f[1] = f[2] = 1;for (int i = 3; i <= 1000000; ++i) {f[i] = (f[i-1] * 2 % mod) + (f[i-2] / 2 % mod);f[i] %= mod;}while (T--) {int n;cin >> n;cout << f[n] << endl;}return 0; }

K. 高度期望

貪心把最小的樹變成最大

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 105; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std;int a[100005]; int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int n, m;cin >> n >> m;int sum = 0;for (int i = 0; i < n; ++i) {cin >> a[i];sum += a[i];}sort(a, a + n);int last = m * n - sum;int ans = 0;for (int i = 0; i < n && last > 0; ++i) {ans++;last -= 1000 - a[i];}cout << ans << endl;return 0; }

L. 最優(yōu)規(guī)劃

并查集先把存在的邊搞到一次，剩下的邊從最小的邊權(quán)開始看是否要加上

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #include <time.h> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) const int maxn = 100001; #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std;int fa[maxn]; int find(int x) {return (x == fa[x]) ? x : fa[x] = find(fa[x]); }int join(int x, int y) {int fx = find(x);int fy = find(y);if (fx == fy) return 0;if (fx > fy) swap(fx, fy);fa[fy] = fx;return 1; }struct ac{LL u, v, d;bool operator < (const ac &t) {return d < t.d;} }a[maxn]; int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0); int n, m, s;cin >> n >> m >> s;for (int i = 1; i <= n; ++i) fa[i] = i;for (int i = 0, u, v, d; i < m; ++i) {cin >> u >> v >> d;join(u, v);}LL ans = 0;for (int i = 0; i < s; ++i) {cin >> a[i].u >> a[i].v >> a[i].d;}sort(a, a + s);for (int i = 0; i < s; ++i) {int u = a[i].u;int v = a[i].v;LL d = a[i].d;int t = join(u, v);if (t) ans += d;}int flag = 1;for (int i = 1; i <= n && flag; ++i) {if (find(i) != find(1)) flag = 0;}if (flag) cout << ans << endl;else cout << "Concubines can't do it.\n";return 0; }

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