文章目录

上海理工大学第二届“联想杯”全国程序设计邀请赛
- A.A-SOUL!
- B.Bheith i ngra le
- C.Counting Cats!
- D.Dahno Dahno
- E.Experiment Class
- I.Identical Day
- J.JXC&Jesus
- K.Kazusa’s Party
- L.Little Witch Academia
- M.Minecraft

上海理工大学第二届“联想杯”全国程序设计邀请赛

A.A-SOUL!

题意：

题解：

代码：

#include <bits/stdc++.h>#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;int const MAXN = 2e5 + 10;
int n, m, T, a[MAXN];
unordered_map<int, int> mp;signed main() {ios_base::sync_with_stdio(false);cin.tie(NULL);cin >> n >> m;for (int i = 1; i <= n; ++i) cin >> a[i];sort(a + 1, a + 1 + n);n = unique(a + 1, a + 1 + n) - a - 1;int res = 0;for (int i = 1; i <= n; ++i) {mp[a[i]] = mp[a[i] - m] + 1;res = max(res, mp[a[i]]);}cout << res;return 0;
}

B.Bheith i ngra le

题意： 给定n * m的网格，对于每一个水平位置i，都有个高度hi。定义“山”为：存在l和r，使得1∼l−11 \sim l-11∼l−1,都有hi<=hi+1hi<=h_{i+1}hi<=hi+1；对于l∼rl\sim rl∼r，都有 hi=hi+1hi=h_{i+1}hi=hi+1；对于 r+1∼nr+1\sim nr+1∼n，都有 hi>=hi+1hi >= h_{i+1}hi>=hi+1。问：给定n和m，求有多少种构成“山”的方案数。1<=n,m<=2000,hi>=01<=n,m<=2000,hi>=01<=n,m<=2000,hi>=0

题解： 组合计数–隔板法。考虑枚举l和r，然后枚举这段区间内的hi。那么得到式子：

∑i=1n∑j=in∑k=1mCi+k−2k−1∗Cn−j+k−1k−1\sum_{i=1}^n\sum_{j=i}^n\sum_{k=1}^m C_{i+k-2}^{k-1}*C_{n-j+k-1}^{k-1}∑i=1n∑j=in∑k=1mCi+k−2k−1∗Cn−j+k−1k−1

解释下这个式子，对于前i-1个位置，要满足hi<=hi+1h_i<=h_{i+1}hi<=hi+1。然后可以选择的高度范围为0∼k−10\sim k-10∼k−1，那么就是在0∼k−10 \sim k-10∼k−1 这k个数字中，选择出 i−1i-1i−1 个数字的方案数。记aia_iai表示选择数字i的次数，那么问题等价于求解方程 a0+a1+...+ak−1=i−1,ai>=0a_0+a_1+...+a_{k-1}=i-1,\ a_i>=0a0+a1+...+ak−1=i−1, ai>=0 的解数目。

因为ai>=0a_i>=0ai>=0，那么按照隔板法往k-1个容器内部加入1个物品，那么就是k−1+i−1=k+i−2k-1+i-1=k+i-2k−1+i−1=k+i−2个物品放入k个箱子里，每个箱子至少一个，即C(k+i-2, k -1)。

如果直接解上面的式子，就是O(n3)O(n^3)O(n3)

由于第一个式子和j没关系，第二个式子和i没关系，那么式子变化为：

∑i=1n∑k=1mCi+k−2k−1∗∑j=inCn−j+k−1k−1\sum_{i=1}^n\sum_{k=1}^m C_{i+k-2}^{k-1}*\sum_{j=i}^nC_{n-j+k-1}^{k-1}∑i=1n∑k=1mCi+k−2k−1∗∑j=inCn−j+k−1k−1

这里记∑j=inCn−j+k−1k−1\sum_{j=i}^nC_{n-j+k-1}^{k-1}∑j=inCn−j+k−1k−1 为 num[i][k]num[i][k]num[i][k]

那么式子转换为：∑i=1n∑k=1mCi+k−2k−1∗num[i][k]\sum_{i=1}^n\sum_{k=1}^m C_{i+k-2}^{k-1} * num[i][k]∑i=1n∑k=1mCi+k−2k−1∗num[i][k]

对于num[i][k]num[i][k]num[i][k] 使用前缀和相减，num[i][k]=pre[n][k]−pre[i−1][k]num[i][k] = pre[n][k] - pre[i-1][k]num[i][k]=pre[n][k]−pre[i−1][k]

代码：

#include <bits/stdc++.h>#define int long long
using namespace std;
int const MAXN = 4200, mod = 1e9 + 7;int fac[MAXN], inv[MAXN];
int n, m, pre[MAXN][MAXN], num[MAXN][MAXN];int cnm(int n, int m) {if (n < m) return 0;if (m == 0) return 1;if (n == m) return 1;return (fac[n] * inv[n - m] % mod) * inv[m] % mod;
}int qmi(int a, int b) {int res = 1;while (b) {if (b & 1) res = res * a % mod;a = a * a % mod;b >>= 1;}return res;
}void init() {inv[1] = 1, fac[1] = 1, fac[0] = 1;for (int i = 2; i < MAXN; i++) fac[i] = fac[i - 1] * i % mod;inv[MAXN - 1] = qmi(fac[MAXN - 1], mod - 2);for (int i = MAXN - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
}signed main() {cin >> n >> m;init();for (int k = 1; k <= m; k++) for (int j = 1; j <= n; j++) {pre[j][k] = pre[j - 1][k] + cnm(n - j + k - 1, k - 1);pre[j][k] %= mod;}for (int k = 1; k <= m; k++) for (int i = 1; i <= n; i++) {num[i][k] = pre[n][k] - pre[i - 1][k];num[i][k] %= mod;}int res = 0;for (int i = 1; i <= n; i++) for (int k = 1; k <= m; k++) {res = (res + cnm(i + k - 2, k - 1) * num[i][k]) % mod;}cout << (res + 1) % mod << endl;return 0;
}

C.Counting Cats!

题意：

题解：

代码：

#include <bits/stdc++.h>using namespace std;int const MAXN = 2e5 + 10;
int n, m, T;
map<string, int> mp;
int main() {ios_base::sync_with_stdio(false);cin.tie(NULL);cin >> n;for (int i = 1; i <= n; i++) {string s;cin >> s;mp[s]++;}int res = mp["cat"] + min(mp["ca"], mp["t"]) + min(mp["c"], mp["at"]);mp["t"] -= min(mp["ca"], mp["t"]);mp["c"] -= min(mp["c"], mp["at"]);res += min({mp["c"], mp["a"], mp["t"]});cout << res;return 0;
}

D.Dahno Dahno

题意：

题解：

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <string>
#include <vector>
#define eps 1.0e-20
using namespace std;
#define mo 1000000000
#define inf 1000000000
#define int long longint cap[505][505];
int node[505], dis[505];
int Stoer_Wagner(int n) {int i, j, now, ans = 0x3fffffff;for (int i = 0; i < n; i++) node[i] = i;while (n > 1) {for (now = 0, i = 1; i < n; i++) dis[node[i]] = 0;for (i = 1; i < n; i++) {swap(node[now], node[i - 1]);for (now = j = i; j < n; j++) {dis[node[j]] += cap[node[i - 1]][node[j]];if (dis[node[now]] < dis[node[j]]) now = j;}}ans = min(ans, dis[node[now]]);for (j = 0; j < n; j++)cap[node[j]][node[now - 1]] = cap[node[now - 1]][node[j]] +=cap[node[j]][node[now]];node[now] = node[--n];}return ans;
}int n, m;
signed main() {cin >> n;int sum = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < n; ++j) {scanf("%lld", &cap[i][j]);sum += cap[i][j];}}int cnt = Stoer_Wagner(n);cout << sum - 2 * cnt << endl;return 0;
}

E.Experiment Class

题意： 给定2条相交直线，给定2个点S和T，现在要求找到一条从S到T的路径，但是路径要经过2条直线各一次。求出最小路径权值。

题解：

初中几何题目

代码：

#include <bits/stdc++.h>using namespace std;
void duicheng(double a, double b, double c, double &x, double &y) {double x1 =((b * b - a * a) * x - 2 * a * b * y - 2 * a * c) / (a * a + b * b);double y1 =((a * a - b * b) * y - 2 * a * b * x - 2 * b * c) / (a * a + b * b);x = x1, y = y1;
}
int main() {double a, b, c, d, x0, y0, x1, y1;scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &x0, &y0, &x1, &y1);if (y0 * x1 > y1 * x0) swap(x0, x1), swap(y0, y1);if (b * c > a * d) swap(a, c), swap(b, d);if (b * x0 < y0 * a) duicheng(-b, a, 0, x0, y0);if (c * y1 < x1 * d) duicheng(-d, c, 0, x1, y1);printf("%.3lf\n", sqrt((x1 - x0) * (x1 - x0) + (y1 - y0) * (y1 - y0)));return 0;
}

I.Identical Day

题意： 给定一个只含0和1构成的字符串，每次修改可以将某个1改为0.问最少修改几次能够使得该字符串的权值小于等于k？字符串的权值和定义为：xxxx见题意 1<=n<=1e5,0<=k<=1e101<=n<=1e5,0<=k<=1e101<=n<=1e5,0<=k<=1e10

题解： 首先看到这个题目第一眼，大家肯定知道平均分最好，比如你操作一次，那么尽可能把剩下的分成2段长度相等的。操作2次，那么尽可能分成3段长度相等的。那么每次到底分几段呢？

因为题目要求操作次数越小越好，那么就可以使用一个优先队列，每次从对头取出多份一段后使得权值变化最大的那个，然后维护变化完的状态，再次放入队列里面。具体见代码：

代码：

#include <bits/stdc++.h>#define int long longusing namespace std;const int maxm = 3e6 + 5;int cal(int x) { return x * (x + 1) / 2; }int get(int all, int cnt) {int num = all / cnt, rem = all % cnt;int c1 = cnt - rem, c2 = cnt - c1;int res = c1 * cal(num) + c2 * cal(num + 1);return res;
}struct Node {int all, cnt;bool operator<(const Node& a) const {int dif1 = get(all, cnt) - get(all - 1, cnt + 1);int dif2 = get(a.all, a.cnt) - get(a.all - 1, a.cnt + 1);return dif1 < dif2;}
};char s[maxm];
int n, k;void solve() {cin >> n >> k;cin >> (s + 1);int now = 0;priority_queue<Node> q;for (int i = 1, sizes = 0; i <= n; i++) {if (s[i] == '1') {sizes++;}if (sizes && (s[i] == '0' || i == n)) {now += get(sizes, 1);q.push({sizes, 1});sizes = 0;}}int res = 0;while (now > k) {Node x = q.top();q.pop();now -= get(x.all, x.cnt);x.all--;x.cnt++;now += get(x.all, x.cnt);if (x.all) {q.push(x);}res++;}cout << res << endl;
}signed main() {ios::sync_with_stdio(0);cin.tie(0);solve();return 0;
}

J.JXC&Jesus

题意：

题解： 通过分析，可以发现其实影响的只有第一个素因子的幂次。然后因为L<=1e7，因此次数最多只有24位。那么直接筛素数的时候找到每个数字的最小素因子，然后暴力幂次即可。

代码：

#include <bits/stdc++.h>#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;inline int read() {int s = 0, w = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();return s * w;
}int const MAXN = 2e7 + 10;
int n, m, T, cnt, L, st[MAXN], prime[MAXN], minp[MAXN];void get_prime(int n) {minp[1] = minp[0] = 1;for (int i = 2; i <= n; ++i) {if (!st[i])prime[cnt++] = i,minp[i] =i;  // 如果这个数字没有被记录，那么这个数字必然为素数，记录一下for (int j = 0; prime[j] <= n / i; ++j) {st[prime[j] * i] = true;  // 筛掉pj*i这个合数minp[prime[j] * i] = prime[j];if (i % prime[j] == 0)break;  // i%pj==0，说明pj是i的最小素因子，因此i*素数的最小素因子也是pj，在i递增的时候也会被筛掉，因此不需要在这里判断}}
}signed main() {// n = read(), m = read(), L = read();scanf("%lld%lld%lld", &n, &m, &L);get_prime(2e7 + 3);// for (int i = 1; i <= 10; ++i) cout << minp[i] << " ";int res = 0;for (int i = L + 1; i <= L + n; ++i) {int pMin = minp[i];int cnt = 0;int now = i;for (int j = 1; j <= 24; j++) {if (now % pMin == 0) {cnt++;now /= pMin;} elsebreak;}cnt /= m;for (int j = 1; j <= cnt; j++) now *= pMin;res += i - now;}printf("%lld\n", res);return 0;
}

K.Kazusa’s Party

题意：

题解：

代码：

#include <bits/stdc++.h>using namespace std;int a[1000], b[1000], n;
signed main() {ios_base::sync_with_stdio(false);cin.tie(NULL);cin >> n;for (int i = 0; i < n; i++) cin >> a[i];for (int i = 0; i < n; i++) cin >> b[i];string ss = "";for (int i = 0; i < n; i++) ss += i + '0';int res = 0;while (1) {int temp = 0;for (int i = 0; i < n; i++) {if (__gcd(a[i], b[ss[i] - '0']) > 1) temp++;}res = max(res, temp);if (!next_permutation(ss.begin(), ss.end())) break;}cout << res << endl;return 0;
}

L.Little Witch Academia

题意：

题解： 先dfs求出所有每行的状态。然后对于h行，设f[i][S]f[i][S]f[i][S]:表示第i行状态为S时的方案数。则转移方程为：f[i][S]=∑S′&S==0f[i−1][S′]f[i][S] = \sum_{S' \& S ==0}f[i-1][S']f[i][S]=∑S′&S==0f[i−1][S′]

这个式子用矩阵快速幂优化一下即可。

**代码： **

// 该代码没有调通，只过了60%样例，大佬懂得麻烦给我教我
#include <bits/stdc++.h>// #define int long long
using namespace std;typedef long long LL;int const N = 128;
int n, mod = 1e9 + 7;
struct mat {int m[N + 10][N + 10];
} unit;void init_unit() {for (int i = 1; i <= N; ++i) unit.m[i][i] = 1;
}void cout_mat(mat x) {for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n; ++j) cout << x.m[i][j] << " ";cout << endl;}cout << endl;
}mat operator*(mat a, mat b) {mat c;memset(c.m, 0, sizeof c.m);  // 记得要初始化为0，否则值不为0for (int i = 1; i <= n; ++i)for (int j = 1; j <= n; ++j)for (int k = 1; k <= n; ++k)c.m[i][j] = (c.m[i][j] + (LL)a.m[i][k] * b.m[k][j] % mod) % mod;return c;
}mat qmi_mat(mat a, int k) {mat res = unit;// cout_mat(a);while (k) {if (k & 1) res = res * a;k >>= 1;a = a * a;}return res;
}mat f1, a;
int T, A, B, W, H;
int mp[N], cnt;void dfs(int cur, int s) {if (cur > W) return void();if (cur == W) return mp[++cnt] = s, void();if (cur + A <= W) dfs(cur + A, (cur + A == W) ? s : s | (1 << (cur + A)));if (cur + B <= W) dfs(cur + B, (cur + B == W) ? s : s | (1 << (cur + B)));
}signed main() {init_unit();cin >> T;for (int i = 1; i <= N; ++i) a.m[1][i] = 1;while (T--) {cnt = 0;cin >> A >> B >> W >> H;dfs(0, 0);n = cnt;// cout << "n := " << n << endl;memset(f1.m, 0, sizeof f1.m);for (int i = 1; i <= n; ++i)for (int j = 1; j <= n; ++j)if (i != j && !(mp[i] & mp[j])) f1.m[i][j] = 1;else f1.m[i][j] = 0;// cout << "f1:\n";// cout_mat(f1);H--;mat res;if (!H) res = unit;else res = qmi_mat(f1, H);// cout << "res:\n";// cout_mat(res);res = a * res;// cout_mat(res);int ans = 0;for (int i = 1; i <= n; ++i) ans = ((LL)ans + res.m[1][i] % mod) % mod;ans = (ans % mod + mod) % mod;cout << ans << endl;}return 0;
}

M.Minecraft

题意：

题解：

代码：

#include <bits/stdc++.h>using namespace std;int const N = 1e5 + 10;
int e[N], ne[N], h[N], idx, d[N];
int n, m, T, H;
char mp[100][100][100];
vector<int> ans;inline int read() {int s = 0, w = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();return s * w;
}set<int> ss;
// 建立邻接表
void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }// 拓扑排序
void top_sort() {priority_queue<int, vector<int>, less<int>> q;  // 这里是求字典序最小的拓扑序，如果求字典序最大的，那么改成// priority_queue<int, vector<int>, less<int>> q;for (int i = 1; i <= 26; ++i)if (!d[i] && ss.count(i)) q.push(i);  // 把入度为0的点加入队列// 当队列不为空时while (q.size()) {auto t = q.top();  // 取队头q.pop();           // 队头出队ans.push_back(t);  // 把这个数字放入答案序列for (int i = h[t]; i != -1; i = ne[i]) {  // 枚举所有队头元素相邻的元素int j = e[i];d[j]--;  // 队头元素出队相当于把与队头元素相连的元素的入度减一if (!d[j]) q.push(j);  // 把入度为0的元素放入队列}}if (ans.size() != ss.size()) {cout << -1 << endl;return;}for (int i = 0; i < ans.size(); ++i) {printf("%c", ans[i] - 1 + 'A');}cout << endl;
}int main() {T = read();while (T--) {ss.clear();memset(h, -1, sizeof h);memset(d, 0, sizeof d);ans.clear();idx = 0;n = read(), m = read(), H = read();for (int i = 0; i < H; ++i) {for (int j = 0; j < n; ++j) {cin >> mp[i][j];for (int k = 0; k < m; ++k) ss.insert(mp[i][j][k] - 'A' + 1);}}for (int i = 0; i < H - 1; ++i) {for (int j = 0; j < n; ++j) {for (int k = 0; k < m; ++k) {if (mp[i + 1][j][k] != mp[i][j][k]) {add(mp[i + 1][j][k] - 'A' + 1, mp[i][j][k] - 'A' + 1);// cout << mp[i + 1][j][k] - 'A' + 1 << "->" << mp[i][j][k] - 'A' + 1 << endl;d[mp[i][j][k] - 'A' + 1]++;}}}}top_sort();}return 0;
}