Codeforces Round #395 (Div. 2) 题解
Posted AC_Arthur
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比赛链接：
本次比赛解决3题（好水呀QAQ）
A. Taymyr is calling you
水题暴力
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF 0xcf
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 100;
int T,n,m,z;
int main()
// ios::sync_with_stdio(false);
// srand((unsigned)time(NULL));
// freopen("out.txt","w",stdout);
// freopen("in.txt","r",stdin);
// scanf("%d",&T);
scanf("%d%d%d", &n, &m, &z);
int ans = 0;
for(int i = 1; i <= z; i++)
if(i % n == 0 && i % m == 0) ++ans;
printf("%d\\n", ans);
return 0;
B. Timofey and cubes
从两边向中间，根据变换次数奇偶来改变
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF 0xcf
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 2e5+10;
int T,n,m,z,a[maxn];
int main()
// ios::sync_with_stdio(false);
// srand((unsigned)time(NULL));
// freopen("out.txt","w",stdout);
// freopen("in.txt","r",stdin);
// scanf("%d",&T);
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
for(int i = 1; i <= n/2; i++)
if(i&1) swap(a[i], a[n-i+1]);
for(int i = 1; i <= n; i++)
printf("%d%c", a[i], i==n?'\\n':' ');
return 0;
C. Timofey and a tree
思路：
随便找一个根做dfs序， 然后建立线段树维护区间最大最小值， 再dfs一遍， 枚举每个点做为根， 那么把线段分成了一些区间， 如果区间最大最小值相等， 那么就只有一种颜色。
细节参见代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF 0xcf
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 2e5+10;
int T,n,m,z,a[maxn],in[maxn],idex = 0, out[maxn], minv[maxn<<2], maxv[maxn<<2];
int ans;
bool ok;
void pushup(int o)
minv[o] = min(minv[o<<1], minv[o<<1|1]);
maxv[o] = max(maxv[o<<1], maxv[o<<1|1]);
void build(int l, int r, int o)
minv[o] = INF;
maxv[o] = -INF;
if(l == r)
minv[o] = maxv[o] = a[l];
return ;
int mid = (l + r) >> 1;
build(l, mid, o<<1);
build(mid+1, r, o<<1|1);
pushup(o);
int querymin(int L, int R, int l, int r, int o)
if(L <= l && r <= R)
return minv[o];
int mid = (l + r) >> 1;
int ans = INF;
if(L <= mid) ans = min(ans, querymin(L, R, l, mid, o<<1));
if(mid < R) ans = min(ans, querymin(L, R, mid+1, r, o<<1|1));
return ans;
int querymax(int L, int R, int l, int r, int o)
if(L <= l && r <= R)
return maxv[o];
int mid = (l + r) >> 1;
int ans = -INF;
if(L <= mid) ans = max(ans, querymax(L, R, l, mid, o<<1));
if(mid < R) ans = max(ans, querymax(L, R, mid+1, r, o<<1|1));
return ans;
vector<int> g[maxn];
int c[maxn];
void dfs(int u, int fa)
int len = g[u].size();
in[u] = ++idex;
a[idex] = c[u];
for(int i = 0; i < len; i++)
int v = g[u][i];
if(v == fa) continue;
dfs(v, u);
out[u] = idex;
void dp(int u, int fa)
int len = g[u].size();
bool flag = trueCodeforces Round #395 (Div. 2) C