链接：Codeforces Round #636 (Div. 3)
A - Candies
题意：求出一个x满足x+2∗x+4∗x+⋯+2k−1∗x=n且k>1
思路：提出x得x∗(1+2+4+⋯+2k−1)=n，从小到大枚举k，直到满足n∣(1+2+4+⋯+2k−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n2个数为偶数，后n2个数为奇数，每个数都为互不相同的正整数，且前n2个数的和等于后n2个数的和
思路：奇数和偶数的差为奇数，如果n2−1为奇数，那么对于前n2−1对奇数和偶数，合起来的差为偶数，那么第n2对奇数和偶数就无法构造出来，所以n2必须为偶数，即n∣4，显然可以构造出这样的数列2 4 6⋯n 1 3 5⋯(n−3) (n+n2−1)
#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∈[1,n2]，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]∈[imin+1,imax+k]，操作两次能使a[i]+a[n-i+1]∈[2,2k]，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+1,2k]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
#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;
}
编写人：刘枝飞