Codeforces Round #636 (Div. 3)
链接：Codeforces Round #636 (Div. 3)
A - Candies
题意：求出一个x满足$x + 2*x + 4*x + \dots + 2^{k-1}*x = n$且$k>1$
思路：提出x得$x*(1+2+4+\dots + 2^{k-1})=n$，从小到大枚举k，直到满足$n \mid (1+2+4+\dots + 2^{k-1})$
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; int T, n; int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%d", &n); int now = 3, t = 4; while (0 != n % now) { now += t; t *= 2; } printf("%d\n", n / now); } return 0; }
B - Balanced Array
题意：给你一个偶数n，要求你构造出一个数列，满足前$\frac{n}{2}$个数为偶数，后$\frac{n}{2}$个数为奇数，每个数都为互不相同的正整数，且前$\frac{n}{2}$个数的和等于后$\frac{n}{2}$个数的和
思路：奇数和偶数的差为奇数，如果$\frac{n}{2}-1$为奇数，那么对于前$\frac{n}{2}-1$对奇数和偶数，合起来的差为偶数，那么第$\frac{n}{2}$对奇数和偶数就无法构造出来，所以$\frac{n}{2}$必须为偶数，即$n\mid 4$，显然可以构造出这样的数列$2\ \ 4\ \ 6\cdots n\ \ 1\ \ 3\ \ 5\cdots (n-3)\ \ (n+\frac{n}{2}-1)$
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; int T, n; int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%d", &n); if (n % 4) { printf("NO\n"); continue; } printf("YES\n"); for (int i = 1; i <= n / 2; i++) printf("%d ", 2 * i); for (int i = 1; i <= n / 2 - 1; i++) printf("%d ", 2 * i - 1); printf("%d\n", n + n / 2 - 1); } return 0; }
C - Alternating Subsequence
题意：就是给你n个数，你需要找出其中的最长子序列，并且这个子序列是正负交替的，并且尽量让这个子序列的和最大
思路：由于要子序列最长，所以把n个数按照正负分成几块(块内元素正负性相同)，然后再在每个块内贪心取一个数即可
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; const int N = 200010; const ll INF = 1000000000000000000; int T, n; ll a[N]; void solve(ll &imax, ll &imin, int l, int r) { for (int i = l; i <= r; i++) { imax = max(imax, a[i]); imin = max(imin, a[i]); } } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%lld", &a[i]); int lst = 1, now = 2; ll res = 0; while (now <= n) { if ((a[now] > 0) != (a[now - 1] > 0)) { ll imax = -INF, imin = -INF; solve(imax, imin, lst, now - 1); if (a[lst] > 0) res += imax; else res += imin; lst = now; } now += 1; } ll imax = -INF, imin = -INF; solve(imax, imin, lst, n); if (a[lst] > 0) res += imax; else res += imin; printf("%lld\n", res); } return 0; }
D - Constant Palindrome Sum
题意：给定一个长度为n的数列，n为偶数，保证每个元素在[1,k]之间，每次操作可以把某个位置的数字变成[1,k]内的任意数字，要求让这个数列满足：对于所有的i$\in$[1,$\frac{n}{2}$]，a[i]+a[n-i+1]是一个定值，问最少的操作次数
思路：设imin=min(a[i],a[n-i+1])，imax=max(a[i],a[n-i+1])，分三种情况，不进行操作a[i]+a[n-i+1]不变，操作一次能够使a[i]+a[n-i+1]$\in$[imin+1,imax+k]，操作两次能使a[i]+a[n-i+1]$\in$[2,2*k]，根据贪心的思想，对于每一对(a[i],a[n-i+1])，[imin+1,imax+k]操作次数加1，[2,imin]和[imax+k+1,2*k]操作次数加2，a[i]+a[n-i+1]操作次数不变，用差分数组维护一下，最后找到[2,2*k]内的最小操作次数即可
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; const int N = 400010; const int INF = 0x3f3f3f3f; int T, n, k, a[N], b[N]; void add(int *d, int l, int r, int x) { d[l] += x; d[r + 1] -= x; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%d%d", &n, &k); for (int i = 1; i <= 2 * k; i++) b[i] = 0; for (int i = 1; i <= n; i++) scanf("%d", &a[i]); for (int i = 1; i <= n / 2; i++) { int imin = min(a[i], a[n - i + 1]); int imax = max(a[i], a[n - i + 1]); if (2 <= imin) add(b, 2, imin, 2); add(b, imin + 1, imax + k, 1); if (imax + 1 <= k) add(b, imax + k + 1, 2 * k, 2); } for (int i = 1; i <= 2 * k; i++) b[i] += b[i - 1]; for (int i = 1; i <= n / 2; i++) b[a[i] + a[n - i + 1]] -= 1; int res = INF; for (int i = 2; i <= 2 * k; i++) res = min(res, b[i]); printf("%d\n", res); } return 0; }
E - Weights Distributing
题意：给定一张n个顶点的有权无向图，m条边和m个权值和三个点a，b，c。问如何分配边权能使a到b，b再到c的权值和最小，求最小值
思路：很直观的感受就是，求出a到b，b再到c的最短路径，然后将m个权值从小到大分配即可，如果同时有多条最短路径，我们应该选择重合最多的两条最短路径(因为这样用的边会更少)，并且此时重合的部分应该分配权值最小的边，所以我们可以用bfs或者dfs求出a，b，c到其他所有点的最短距离da[],db[],dc[]，然后枚举每个点x，假设a到b，b再到c的路径拆为a到x，x到b，b到x，x再到c，枚举所有的点x，一定会包含上述的两种(两条路径重叠与不重叠)的情况，并且此时b到x应该分配最小的边权，将m个边权排序，求前缀和sum[]，那么对于点x，答案就是sum[db[x]]+sum[da[x]+db[x]+dc[x]]
参考：https://blog.csdn.net/qq_34261446/article/details/105753266
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <queue> using namespace std; typedef long long ll; const int N = 200010; const int INF = 0x3f3f3f3f; struct node { int to, nex; }; int T, n, m, a, b, c; ll p[N], sum[N]; int head[N], cnt, da[N], db[N], dc[N]; node edge[2 * N]; queue<int> q; void init() { cnt = 0; for (int i = 1; i <= n; i++) head[i] = 0; for (int i = 1; i <= n; i++) da[i] = db[i] = dc[i] = INF; } void add_edge(int u, int v) { edge[++cnt].to = v; edge[cnt].nex = head[u]; head[u] = cnt; } void bfs(int s, int *d) { while (!q.empty()) q.pop(); d[s] = 0; q.push(s); while (!q.empty()) { int u = q.front(); q.pop(); for (int i = head[u]; 0 != i; i = edge[i].nex) { int v = edge[i].to; if (d[v] != INF) continue; d[v] = d[u] + 1; q.push(v); } } } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &T); while (T--) { scanf("%d%d%d%d%d", &n, &m, &a, &b, &c); init(); for (int i = 1; i <= m; i++) scanf("%lld", &p[i]); sort(p + 1, p + m + 1); for (int i = 1; i <= m; i++) sum[i] = sum[i - 1] + p[i]; for (int i = 1; i <= m; i++) { int u, v; scanf("%d%d", &u, &v); add_edge(u, v); add_edge(v, u); } bfs(a, da); bfs(b, db); bfs(c, dc); ll res = 1000000000000000000; for (int i = 1; i <= n; i++) { if (da[i] + db[i] + dc[i] > m) continue; res = min(res, sum[db[i]] + sum[da[i] + db[i] + dc[i]]); } printf("%lld\n", res); } return 0; }
