【AtCoder】AGC024

时间  2019-12-10
标签 AtCoder agc024 agc 繁體版
原文   原文链接

A - Fairness

若是奇数次是b - a
不然是a - bnode

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
//#define ivorysi
using namespace std;
typedef long long int64;
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 A,B,C,K;
void Solve() {
    read(A);read(B);read(C);read(K);
    if(K & 1) {out(B - A);enter;}
    else {out(A - B);enter;}
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

B - Backfront

找数值最长的连续的一段子序列，而后将剩下的数必需要移动了ios

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
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,a[MAXN],pos[MAXN],dp[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
    read(a[i]);
    pos[a[i]] = i;
    }
    dp[1] = 1;
    for(int i = 2 ; i <= N ; ++i) {
    if(pos[i - 1] < pos[i]) dp[i] = dp[i - 1] + 1;
    else dp[i] = 1;
    }
    int res = N;
    for(int i = 1 ; i <= N ; ++i) {
    res = min(res,N - dp[i]);
    }
    out(res);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

C - Sequence Growing Easy

数列确定是一段连续递增的好几段拼起来，都找到加上最大值就能够了c++

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
//#define ivorysi
#define pb push_back
#define MAXN 200005
#define eps 1e-12
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define fi first
#define se second
#define mo 974711
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 - '0' + c;
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) putchar('-'),x = -x;
    while(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,a[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    a[0] = -1;
    for(int i = 1 ; i <= N ; ++i) {
    if(a[i - 1] < a[i] - 1) {
        puts("-1");
        return;
    }
    }
    int64 ans = 0;
    for(int i = N ; i >= 1 ; --i) {
    if(a[i] == a[i + 1] - 1) continue;
    ans += a[i];
    }
    printf("%lld\n",ans);
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Isomorphism Freak

答案是直径的长度 / 2 + 1spa

咱们对于直径长度为奇数，也就是有偶数个点，若是直径长度增长颜色也要增长，因此咱们不增长直径
咱们枚举做为直径中心的两个点，咱们能达到这个要求仅当距离中心距离相同的点点度相同便可rest

对于直径长度为偶数，也就是有奇数个点，咱们分两种状况，第一种是只有一个中心
第二种是有两个中心，咱们枚举两个中心，而后要求这两个中心所在的树的深度最大的不超过中心的一半便可code

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 105
//#define ivorysi
using namespace std;
typedef long long int64;
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);
}
struct node {
    int to,next;
}E[MAXN * 2];
int N,sumE,head[MAXN],A[MAXN],B[MAXN],dep[MAXN],ch[MAXN],x[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;E[sumE].next = head[u];head[u] = sumE;
}
int Calc(int u,int fa) {
    int res = dep[u];
    ch[u] = 0;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        ch[u]++;
        dep[v] = dep[u] + 1;
        res = max(res,Calc(v,u));
    }
    }
    return res;
}
void Solve() {
    read(N);
    if(N == 1) {
    puts("1 2");return;
    }
    for(int i = 1 ; i < N ; ++i) {
    read(A[i]);read(B[i]);
    add(A[i],B[i]);add(B[i],A[i]);
    }
    int D = 0;
    for(int i = 1 ; i < N ; ++i) {
    dep[A[i]] = dep[B[i]] = 0;
    D = max(D,Calc(A[i],B[i]) + Calc(B[i],A[i]) + 1);
    }
    out(D / 2 + 1);space;
    if(D & 1) {
    int64 ans = 9e18;
    for(int i = 1 ; i < N ; ++i) {
        dep[A[i]] = dep[B[i]] = 0;
        int64 tmp = 2;
        if(Calc(A[i],B[i]) == D / 2 && Calc(B[i],A[i]) == D / 2) {
        memset(x,0,sizeof(x));
        for(int j = 1 ; j <= N ; ++j) x[dep[j]] = max(x[dep[j]],ch[j]);
        for(int k = 0 ; k < D / 2 ; ++k) tmp = tmp * x[k];
        ans = min(ans,tmp);
        }
        
    }
    out(ans);enter;
    }
    else {
    int64 ans = 9e18;
    for(int i = 1 ; i < N ; ++i) {
        dep[A[i]] = dep[B[i]] = 0;
        int64 tmp = 1;
        int x0 = Calc(A[i],B[i]),x1 = Calc(B[i],A[i]);
        if(x0 > x1) {swap(x0,x1),swap(A[i],B[i]);}
        if(x0 == D / 2 - 1 && x1 == D / 2) {
        dep[B[i]] = 0;
        Calc(B[i],0);
        memset(x,0,sizeof(x));
        for(int j = 1 ; j <= N ; ++j) x[dep[j]] = max(x[dep[j]],ch[j]);
        for(int k = 0 ; k < D / 2 ; ++k) tmp = tmp * x[k];
        ans = min(ans,tmp);
        }
        
    }
    
    for(int i = 1 ; i < N ; ++i) {
        dep[A[i]] = dep[B[i]] = 0;
        int x0 = Calc(A[i],B[i]),x1 = Calc(B[i],A[i]);
        if(max(x0,x1) <= D / 2) {
        memset(x,0,sizeof(x));
        int64 tmp = 2;
        for(int j = 1 ; j <= N ; ++j) x[dep[j]] = max(x[dep[j]],ch[j]);
        for(int k = 0 ; k <= D ; ++k) {
            if(x[k]) tmp *= x[k];
        }
        ans = min(ans,tmp);
        }
    }
    
    out(ans);enter;
    }
    
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Sequence Growing Hard

Let's restate the problem = =get

咱们就至关于先有一个一个标号为0，值为0的点string

而后每次新加一个标号为k的点在p点上，咱们要新加的k点严格大于p点的值it

这个能够用dp来完成io

dp[u][x]表示用u个点，根节点值为x的方案数，标号就是0 - u - 1
答案就是dp[N + 1][0]

转移就是
\(dp[n] = \sum_{y > x} dp[n - k][x] * dp[k][y] \binom{n - 2}{k - 1}\)
咱们每次把y当作1号点，x当作0号点能够不重不漏

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
//#define ivorysi
using namespace std;
typedef long long int64;
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,K,MOD;
int dp[305][305],sum[305][305],C[305][305];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);read(K);read(MOD);
    C[0][0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
    C[i][0] = 1;
    for(int j = 1 ; j <= i ; ++j) {
        C[i][j] = inc(C[i - 1][j - 1],C[i - 1][j]);
    }
    }
    for(int j = 0 ; j <= K ; ++j) {sum[1][j] = (j + 1) % MOD;dp[1][j] = 1;}
    for(int i = 2 ; i <= N + 1; ++i) {
    for(int j = 1 ; j < i ; ++j) {
        for(int k = 0 ; k <= K ; ++k) {
        update(dp[i][k],mul(mul(dp[j][k],inc(sum[i - j][K],MOD - sum[i - j][k])),C[i - 2][i - j - 1]));
        }
    }
    sum[i][0] = dp[i][0];
    for(int k = 1 ; k <= K ; ++k) sum[i][k] = inc(sum[i][k - 1],dp[i][k]);
    }
    out(dp[N + 1][0]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Simple Subsequence Problem

咱们只要构造出一张图
例如110[00011001]
这个点能够转移到
1100[0011001]和1101[1001]和110[]
咱们若s在集合中[s] = 1
每一个点t的值是t[]
若是t是s的子序列，转移的路径是惟一的，因此只要用dp求路径条数便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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){putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}

int N,K,cnt;
char s[25][(1 << 20) + 5];
vector<int> v[25];
int dp[42000005];
int p0[25],p1[25];
void Solve() {
    read(N);read(K);
    for(int i = 0 ; i <= N ; ++i) scanf("%s",s[i]);
    for(int i = 1 ; i <= N ; ++i) {
        v[i].resize((i + 1) * (1 << i));
        int s = v[i].size();
        for(int j = 0 ; j < s ; ++j) v[i][j] = ++cnt;
    }
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 0 ; j < (1 << i) ; ++j) {
            if(s[i][j] == '1') {
                dp[v[i][j * (i + 1) + i]]++;
            }
        }
    }
    for(int i = N ; i >= 1 ; --i) {
        for(int S = 0 ; S < (1 << i) ; ++S) {
            p0[0] = -1,p1[0] = -1;
            for(int j = 0 ; j <= i ; ++j) {
                p1[j + 1] = p1[j];p0[j + 1] = p0[j];
                if(S >> j & 1) p1[j + 1] = j;
                else p0[j + 1] = j;
            }
            for(int j = i ; j >= 0 ; --j) {
                int u = v[i][S * (i + 1) + j];
                int b = S & ((1 << j) - 1),f = S >> j;
                if(dp[u]) {
                    if(p0[j] != -1) {
                        int p = ((f << 1) << p0[j]) + (S & (1 << p0[j]) - 1);
                        int l = p0[j] + 1 + i - j;
                        dp[v[l][p * (l + 1) + p0[j]]] += dp[u];
                    }
                    if(p1[j] != -1) {
                        int p = ((f << 1 | 1) << p1[j]) + (S & (1 << p1[j]) - 1);
                        int l = p1[j] + 1 + i - j;
                        dp[v[l][p * (l + 1) + p1[j]]] += dp[u];
                    }
                    if(j != 0 && j != i) {
                        int l = i - j;
                        dp[v[l][f * (l + 1)]] += dp[u];
                    }
                }
            }
        }
    }
    for(int i = N ; i >= 1 ; --i) {
        for(int j = 0 ; j < (1 << i) ; ++j) {
            if(dp[v[i][j * (i + 1)]] >= K) {
                for(int k = i - 1; k >= 0 ; --k) {
                    putchar('0' + ((j >> k) & 1));
                }
                enter;
                return;
            }
        }
    }
    puts("");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}
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