【AtCoder】ARC088

C - Multiple Gift

題解

首項是X，每次乘個2，暴力統計

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 2005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int64 X,Y;

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(X);read(Y);
    int l = 1;
    while(X <= Y / 2) {++l;X *= 2;}
    out(l);enter;
}

D - Wide Flip

題解

從大到小枚舉K（也能夠二分）
若是\(2K <= N\)，此時\(K\)必定合法
不然看前\(K\)個和後\(K\)個交出來的中間一段，這段的前面和後面1和0是能夠單個位置隨便改的（用一次K + 1的區間再用一次K的區間）
只要看中間一段是否是全1或者全0便可

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 2005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
char s[1000005];
int sum[100005],N;
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    scanf("%s",s + 1);
    N = strlen(s + 1);
    for(int i = 1 ; i <= N ; ++i) {
    sum[i] = sum[i - 1];
    if(s[i] == '1') sum[i]++;
    }
    for(int K = N ; K >= 1 ; --K) {
    int r = K;
    int l = N - K + 1;
    if(l > r) {
        out(K);enter;return 0;
    }
    if(sum[r] - sum[l - 1] == 0 || sum[r] - sum[l - 1] == r - l + 1) {
        out(K);enter;return 0;
    }
    }
}

E - Papple Sort

題解

只要每次把靠邊的一個字符的匹配點找出來一樣後移而後一塊兒刪除便可
就是討論一下
...A...A...B...B..
A要和這個一對B交換兩次
B若是要過去也要換兩次，因此都同樣
A..B..A..B
A要和B換一次，若是B到外面也是和A換一次，因此都同樣
樹狀數組維護一下前綴有多少數便可

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
char s[MAXN];
int N;
vector<int> c[30];
bool vis[MAXN];
int tr[MAXN];
void Init() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
}
int lowbit(int x) {return x & -x;}
void Insert(int x,int v) {
    while(x <= N) {
    tr[x] += v;
    x += lowbit(x);
    }
}
int Query(int x) {
    int res = 0;
    while(x > 0) {
    res += tr[x];
    x -= lowbit(x);
    }
    return res;
} 
void Solve() {
    int cnt = 0;
    for(int i = N ; i >= 1 ; --i) {
    c[s[i] - 'a'].pb(i);
    }
    for(int i = 0 ; i < 26 ; ++i) {
    if(c[i].size() & 1) ++cnt;
    }
    if(cnt >= 2) {puts("-1");return;}
    for(int i = 1 ; i <= N ; ++i) Insert(i,1);
    int64 ans = 0;
    int k = 0;
    for(int i = N ; i >= 1 ; --i) {
    if(vis[i]) continue;
    int t = c[s[i] - 'a'].back();c[s[i] - 'a'].pop_back();
    if(t == i){ ++k;continue;}
    ans += Query(t - 1) + k;
    vis[t] = 1;
    Insert(t,-1);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}

F - Christmas Tree

題解

啊這個只要dp一下第一個值再二分。。。二分判的方法是……
woc這不NOIPD1T3啊怎麼如出一轍啊

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
int dp[MAXN],A,lim,res;
int val[MAXN],tot;
bool vis[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void Init() {
    read(N);
    int a,b;
    for(int i = 1 ; i < N ; ++i) {read(a);read(b);add(a,b);add(b,a);}
    
}
void dfs(int u,int fa) {
    int son = 0;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        dfs(v,u);
        ++son;
        dp[u] += dp[v];
    }
    }
    dp[u] -= son / 2;
    if(son % 2 == 0 && fa) dp[u]++;
}
struct BIT {
    int tr[MAXN],s;
    void clear() {
    for(int i = 1 ; i <= s ; ++i) tr[i] = 0;
    }
    int lowbit(int x) {return x & -x;}
    int query(int x) {
    int res = 0;
    while(x > 0) {
        res += tr[x];
        x -= lowbit(x);
    }
    return res;
    }
    void insert(int x,int v) {
    while(x <= s) {
        tr[x] += v;
        x += lowbit(x);
    }
    }
    int find_max() {
    int L = 1,R = s;
    int x = query(s);
    if(x == 0) return 0;
    while(L < R) {
        int m = (L + R + 1) >> 1;
        if(query(m - 1) != x) L = m;
        else R = m - 1;
    }
    return L;
    }
}BST;
void calc(int u,int fa) {
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        calc(v,u);
    }
    }
    tot = 0;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        val[++tot] = dp[v];
    }
    }
    if(!tot) {dp[u] = 1;++res;return;}
    BST.clear();
    sort(val + 1,val + tot + 1);
    for(int i = 1 ; i <= tot ; ++i) vis[i] = 1;
    BST.s = tot;int p = 0;
    for(int i = tot ; i >= 1 ; --i) {
    if(!vis[i]) continue;
    while(p < i && val[p + 1] + val[i] <= lim) {
        ++p;BST.insert(p,1);
    }
    if(p >= i) BST.insert(i,-1);
    int t = BST.find_max();
    if(t) {vis[t] = 0;vis[i] = 0;BST.insert(t,-1);--res;}
    }
    if(fa) {
    dp[u] = 1;++res;
    for(int i = 1 ; i <= tot ; ++i) {
        if(vis[i]) {
        if(val[i] + 1 > lim) break;
        dp[u] += val[i];--res;
        break;
        }
    }
    }
}
bool check(int m) {
    lim = m;res = 0;
    calc(1,0);
    return res <= A;
}
void Solve() {
    dfs(1,0);
    out(dp[1]);space;
    A = dp[1];
    int L = 1,R = N - 1;
    while(L < R) {
    int mid = (L + R) >> 1;
    if(check(mid)) R = mid;
    else L = mid + 1;
    }
    out(R);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}