2020 HDU Multi-University Training Contest 6（部分)

6827 Road To The 3rd Building

算每个sks_ksk的贡献，s1s_1s1要乘的分子和为(1+1/2+…+1/n)，s2s_2s2为(1+2/2+…+1/n)，容易发现规律s2s_2s2要乘以的分子和为s1s_1s1要乘的分子和+(1/2+…+1/(n-1))，依次类推，每次分子和是上一个加上sum[st-1]-sum[ed-1-1]（sum[i]表示1+1/2+…+1/i的逆元)。
此外容易发现sis_isi与sn−i+1s_{n-i+1}sn−i+1要乘的分子和是相同的。
因此可以O(n)来计算答案。

#include<cstdio>
using namespace std;
typedef long long ll;
const ll p = 1e9 + 7;
const int maxn = 2e5 + 5;
int T, n;
ll s[maxn], inv[maxn], sum[maxn];
int main(void) {//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);inv[0] = inv[1] = 1;for (int i = 2; i < maxn; i++) {inv[i] = (p - p / i) * inv[p % i] % p;}for (int i = 1; i < maxn; i++)sum[i] = (sum[i - 1] + inv[i]) % p;scanf("%d", &T);while (T--) {scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%lld", &s[i]);ll ans = 0, invSum = 0;int m = n / 2, st = 1, ed = n;for (int i = 1; i <= m; i++) {invSum = (invSum + ((sum[ed] - sum[st - 1]) % p + p) % p) % p;ans = (ans + (s[i] + s[n - i + 1]) % p * invSum % p) % p;st++; ed--;}if (n & 1) {m++; invSum = (invSum + ((sum[ed] - sum[st - 1]) % p + p) % p) % p;ans = (ans + s[m] * invSum % p) % p;}printf("%lld\n", ans * inv[n] % p * inv[n + 1] % p * 2 % p);}return 0;
}

6828 Little Rabbit’s Equation

Little Rabbit’s Equation
形式为：非负数运算符非负数 = 非负数
注意：最小为2进制，开long long

#include<cstdio>
#include<iostream>
#include<string>
#include<stack>
using namespace std;
typedef long long ll;
int num[256];
string line;
ll cal(string s, int digit) {int pos = s.length();char ch = '+';for (int i = 0; i < s.length(); i++) {if (isdigit(s[i]) || isalpha(s[i]))continue;ch = s[i];pos = i;break;}ll t1 = 0, t2 = 0;for (int i = 0; i < pos; i++) {if (isdigit(s[i]))t1 = t1 * digit + s[i] - '0';if (isalpha(s[i]))t1 = t1 * digit + num[s[i]];}for (int i = pos + 1; i < s.length(); i++) {if (isdigit(s[i]))t2 = t2 * digit + s[i] - '0';if (isalpha(s[i]))t2 = t2 * digit + num[s[i]];}ll res = t1;if (ch == '+')res = res + t2;if (ch == '-')res = res - t2;if (ch == '*')res = res * t2;if (ch == '/') {if (res % t2 == 0) res = res / t2;else res = -1;}return res;
}
int main(void) {num['A'] = 10; num['B'] = 11; num['C'] = 12;num['D'] = 13; num['E'] = 14; num['F'] = 15;while (cin >> line) {int minxx = 0;for (int i = 0; i < line.length(); i++) {if (isdigit(line[i])) {if (minxx<9 && line[i] - '0' > minxx)minxx = line[i] - '0';}else if (isalpha(line[i])) {if (num[line[i]] > minxx)minxx = num[line[i]];}}int ans = -1, pos = line.find('=');for (int digit = minxx + 1; digit <= 16; digit++) {if (ans != -1)break;ll a = cal(line.substr(0, pos), digit);ll b = cal(line.substr(pos + 1), digit);if (a == b)ans = digit;}if (ans == 1)ans++;printf("%d\n", ans);}return 0;
}

6829 Borrow

对于数x,y,z(x≥y≥z)x,y,z(x\geq y \geq z)x,y,z(x≥y≥z)，若和不为3的倍数显然无解输出-1；取mmm作为平均数，可先从xxx中拿出x−mx-mx−m元钱分配给yyy和zzz，假设分配给zzz为ttt元，则有序列，z+t,m,y+x−m−tz+t,m,y+x-m-tz+t,m,y+x−m−t，可以看成是m−c,m,m+cm-c,m,m+cm−c,m,m+c的等差序列，然后又要从m+cm+cm+c中拿ccc元分配给另外两人，假设期望值为fcf_cfc，则得到公式fc=c+12c∑i=0cCcifif_c=c+\frac{1}{2^c}\sum^c_{i=0}C_c^if_ifc=c+2c1∑i=0cCcifi
摘自：https://www.cnblogs.com/wasa855/p/13448251.html

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod = 998244353;
const ll maxn = 1e6 + 5;
ll t, a[3];
ll qpow(ll a, ll b) {ll res = 1;while (b){if (b & 1)res = (res * a) % mod;a = a * a % mod;b >>= 1;}return res;
}
ll Inv(ll x) { return qpow(x, mod - 2); }
ll inv[maxn], pow2[maxn], fac[maxn];
ll Cmn(ll m, ll n) {if (!n)return 1;return fac[m] * inv[n] % mod * inv[m - n] % mod;
}
int main(void) {inv[0] = inv[1] = pow2[0] = fac[0] = fac[1] = 1, pow2[1] = 2;for (ll i = 2; i < maxn; i++) {inv[i] = inv[i - 1] * Inv(i) % mod;pow2[i] = pow2[i - 1] * 2 % mod;fac[i] = fac[i - 1] * i % mod;}scanf("%lld", &t);while (t--){scanf("%lld %lld %lld", &a[0], &a[1], &a[2]);if ((a[0] + a[1] + a[2]) % 3) { printf("-1\n"); continue; }ll m = (a[0] + a[1] + a[2]) / 3;sort(a, a + 3);ll ans = 0;for (ll t = 0; t <= a[2] - m; t++) {ll now = min(a[0] + t, a[1] + a[2] - m - t);ll c = m - now;ans = (ans + Cmn(a[2] - m, t) * 2 % mod * c % mod) % mod;}printf("%lld\n", (ans * Inv(pow2[a[2] - m]) % mod + a[2] - m) % mod);}return 0;
}

6830 Asteroid in Love

遍历两个类型点组成线段，与第三类的点构成三角形，易知第三类的点一定是凸包上的点才有面积最大的效果。对凸包分为上下两部分，进行三分求极限。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
const int maxn = 3e3 + 5;
const double eps = 1e-8;
typedef long long ll;
int sgn(double x) {if (fabs(x) < eps)return 0;else return x < 0 ? -1 : 1;
}
struct Point {ll x, y;Point() {}Point(ll x, ll y) :x(x), y(y) {}Point operator + (Point B) { return Point(x + B.x, y + B.y); }Point operator - (Point B) { return Point(x - B.x, y - B.y); }bool operator < (Point B) {return sgn(x - B.x) < 0 || (sgn(x - B.x) == 0 && sgn(y - B.y) < 0);}
};
typedef Point Vector;
ll Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; }
void Convex_hull(Point* p, int n, Point* ch, int& top, int op) {if (op) {for (int i = 0; i < n; i++) {while (top > 1 && sgn(Cross(ch[top - 1] - ch[top - 2], p[i] - ch[top - 2])) <= 0)top--;ch[top++] = p[i];}}else {for (int i = n - 1; i >= 0; i--) {while (top > 1 && sgn(Cross(ch[top - 1] - ch[top - 2], p[i] - ch[top - 2])) <= 0)top--;ch[top++] = p[i];}}
}int T, n, cnt[3];
Point p[3][maxn], ch1[maxn], ch2[maxn];ll getArea(Point a, Point b, int top1, int top2) {ll res = 0;int l = 0, r = top1;while (r - l >= 3) {int lmid = (l + l + r) / 3;int rmid = (l + r + r) / 3;if (abs(Cross(a - b, a - ch1[lmid])) > abs(Cross(a - b, a - ch1[rmid])))//比较面积大小r = rmid;elsel = lmid;}for (int i = l; i <= r; i++)res = max(res, abs(Cross(a - b, a - ch1[i])));l = 0, r = top2;while (r - l >= 3) {int lmid = (l + l + r) / 3;int rmid = (l + r + r) / 3;if (abs(Cross(a - b, a - ch2[lmid])) > abs(Cross(a - b, a - ch2[rmid])))r = rmid;elsel = lmid;}for (int i = l; i <= r; i++)res = max(res, abs(Cross(a - b, a - ch2[i])));return res;
}int main(void) {//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);scanf("%d", &T);while (T--) {scanf("%d", &n);cnt[0] = cnt[1] = cnt[2] = 0;for (int i = 0; i < n; i++) {double a, b; int c;scanf("%lf %lf %d", &a, &b, &c);p[c][cnt[c]++] = Point(a, b);}sort(p[2], p[2] + cnt[2]);int top1 = 0, top2 = 0;Convex_hull(p[2], cnt[2], ch1, top1, 1);Convex_hull(p[2], cnt[2], ch2, top2, 0);ll ans = 0;for (int i = 0; i < cnt[0]; i++) {for (int j = 0; j < cnt[1]; j++) {ans = max(ans, getArea(p[0][i], p[1][j], top1 - 1, top2 - 1));}}if (ans & 1)printf("%lld.5\n", ans / 2);else printf("%lld.0\n", ans / 2);}return 0;
}

6831 Fragrant numbers

区间DP题，预处理，1e8左右的操作数时间挺够的。

#include<cstdio>
using namespace std;
const int maxn = 5000 + 5;
const int maxl = 14;
bool dp[15][15][maxn];
int num[15] = { 0,1,1,4,5,1,4,1,9,1,9,1,1,4,5 };
int N;
int main(void) {for (int i = 1; i <= 14; i++) {int x = 0;for (int j = i; j <= maxl; j++) {x = x * 10 + num[j];if (x >= maxn)break;dp[i][j][x] = 1;}}for (int len = 2; len < maxl; len++)for (int L = 1; L + len - 1 <= maxl; L++) {int R = L + len - 1;for (int mid = L; mid < R; mid++)for (int val1 = 1; val1 < maxn; val1++)if (dp[L][mid][val1])for (int val2 = 1; val2 < maxn; val2++)if (dp[mid + 1][R][val2]) {if (val1 + val2 < maxn)dp[L][R][val1 + val2] = 1;if (val1 * val2 < maxn)dp[L][R][val1 * val2] = 1;}}//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int t;scanf("%d", &t);while (t--) {scanf("%d", &N);int ans = -1;for (int i = 1; i <= 14; i++)if (dp[1][i][N]) { ans = i; break; }printf("%d\n", ans);}return 0;
}

6832 A Very Easy Graph Problem

显然新加入的边一定比已经生成的路径都要大，所以建立最小生成树就能实现所求的最小距离。一次性DFS求出每条边的一个方向上的所有黑点和白点，则相反方向的黑点白点可以由总数减去已求的部分。交叉相乘相加再乘边权就是该边对答案的贡献。

注意：在使用权值时最好临时获取对应的2^i次方，即开始只存边的编号，否则可能出现数据错误(原因未知)

#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
const ll p = 1e9 + 7;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 5;
int T, n, m, a[maxn], fa[maxn], vis[maxn];
ll white[maxm], black[maxm], w[maxm];
struct cosId {int to, id;cosId(int to, int id) :to(to), id(id) {}
};
vector<cosId>e[maxn];
struct edge {int from, to;ll weight;edge(int u, int v, ll w) :from(u), to(v), weight(w) {}bool operator<(const edge& rhs)const {return weight > rhs.weight;}
};
struct Res {int wtn, bkn;Res(int w = 0, int b = 0) :wtn(w), bkn(b) {}
};
void init() {for (int i = 1; i <= n; i++) {fa[i] = i;e[i].clear();}memset(white, 0, sizeof(white));memset(black, 0, sizeof(black));memset(vis, 0, sizeof(vis));
}
int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y) {x = find(x);y = find(y);if (x != y)fa[y] = x;
}
Res dfs(int u) {vis[u] = 1;Res res;for (int i = 0; i < e[u].size(); i++) {int v = e[u][i].to;if (vis[v])continue;Res tmp = dfs(v);white[e[u][i].id] = tmp.wtn;black[e[u][i].id] = tmp.bkn;res.wtn += tmp.wtn;res.bkn += tmp.bkn;}if (a[u])res.bkn++;else res.wtn++;return res;
}
int main(void) {//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);w[0] = 1;for (int i = 1; i < maxm; i++)w[i] = (w[i - 1] << 1) % p;scanf("%d", &T);while (T--) {scanf("%d %d", &n, &m); init();ll Bsum = 0, Wsum = 0;for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);if (a[i])Bsum++;else Wsum++;}priority_queue<edge>pq;for (int i = 1; i <= m; i++) {int u, v;scanf("%d %d", &u, &v);pq.push(edge(u, v, i));}vector<int>ans;int cnt = 1;while (cnt != n) {int u, v;u = pq.top().from;v = pq.top().to;if (find(u) == find(v)) {pq.pop(); continue;}merge(u, v);e[u].push_back(cosId(v, cnt - 1));e[v].push_back(cosId(u, cnt - 1));ans.push_back(pq.top().weight);pq.pop();cnt++;}dfs(1);ll sum = 0;for (int i = 0; i < ans.size(); i++) {ll tmpB = Bsum - black[i];ll tmpW = Wsum - white[i];sum = ((sum + (tmpB * white[i] % p *w[ans[i]] % p) % p + (tmpW * black[i] % p * w[ans[i]] % p))) % p;}printf("%lld\n", sum);}return 0;
}

6835 Divisibility

题意是f(f(…f(y)…))被x整除等价于y被x整除，显然x>=b时不成立，式子最后的结果小b

若x<b，当f(f(…f(y)…))被x整除时有：
y=c1∗(bn−1−1)+c2∗(bn−2−1)+⋯+cn∗(b0−1)+(c1+c2+⋯+cn)y=c_1*(b^{n-1}-1)+c_2*(b^{n-2}-1)+\cdots+c_n*(b^{0}-1)+(c_1+c_2+\cdots+c_n)y=c1∗(bn−1−1)+c2∗(bn−2−1)+⋯+cn∗(b0−1)+(c1+c2+⋯+cn)
最后的部分被x整除，若y被x整除，则b-1为x的倍数即可，容易验证，b-1不为x的倍数时，y不被x整除。

#include<cstdio>
using namespace std;
typedef long long ll;
int t;
ll b, x;
int main(void) {scanf("%d", &t);while (t--){scanf("%lld%lld", &b, &x);printf("%c\n", (b - 1) % x == 0 ? 'T' : 'F');}return 0;
}

6836 Expectation

矩阵树定理用于无向图：
n个点的图，建立n*n矩阵，矩阵中第i行第i列的元素为结点i的度数，第i行第j列为结点i与结点j的边数的负值，任意去掉一行和一列后的行列式值即为改图的生成树个数。

如何证明未知，详细请去百度。

#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
typedef pair<int, int>PII;
const ll mod = 998244353;
const int maxm = 1e4 + 5;
const int maxn = 105;
int t, n, m;
ll qpow(ll a, ll b) {ll res = 1;while (b) {if (b & 1)res = (res * a) % mod;a = a * a % mod;b >>= 1;}return res;
}
ll inv(ll x) { return qpow(x, mod - 2); }
struct edge {int u, v;ll w;
}e[maxm];
ll K[maxn][maxn];
ll Guass() {ll res = 1;for (int i = 1; i < n; i++) {//依次把第i列，从第i+1行到第n行变为0for (int j = i + 1; j < n; j++) {while (K[j][i]) {//直到为0ll t = K[i][i] / K[j][i];//计算第i列,第i行是第j行的几倍for (int k = i; k < n; k++) {K[i][k] = (K[i][k] - t * K[j][k] % mod + mod) % mod;swap(K[i][k], K[j][k]);//把第i行元素变为0后与第j行交换}res = -res;//每次交换行 值变为原来的负数}}if (!K[i][i])return 0;res = (res * K[i][i] % mod + mod) % mod;//将行列式化为上三角行列式后 值为对角线元素的乘积}return res;
}
int main(void) {//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);scanf("%d", &t);while (t--) {scanf("%d %d", &n, &m);memset(K, 0, sizeof(K));for (int i = 0; i < m; i++) {int u, v; ll w;scanf("%d %d %lld", &u, &v, &w);e[i].u = u; e[i].v = v; e[i].w = w;K[u][v]--;K[v][u]--;K[u][u]++;K[v][v]++;}ll sum = Guass(), ans = 0;for (int k = 0; k <= 30; k++) {memset(K, 0, sizeof(K));for (int i = 0; i < m; i++) {int u = e[i].u, v = e[i].v;ll ee = (e[i].w & (1LL << k)) ? 1 : 0;K[u][u] += ee;K[v][v] += ee;K[u][v] -= ee;K[v][u] -= ee;}ans = (ans + Guass() * (1LL << k) % mod) % mod;}printf("%lld\n", (ans * inv(sum) % mod + mod) % mod);}return 0;
}