【Atcoder】ARC102 题解

C - Triangular Relationship

题解

枚举一个数%K的值然后统计另两个

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 200005
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
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) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int N,K;
int cnt[200005];
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);read(K);
    int64 ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
        cnt[i % K]++;
    }
    for(int i = 0 ; i < K ; ++i) {
        if((K - i) * 2 % K == 0) {
            ans += 1LL * cnt[i] * cnt[(K - i) % K] * cnt[(K - i) % K];
        }
    }
    out(ans);enter;
    return 0;
}

D - All Your Paths are Different Lengths

题解

感觉自己万分智障，没切掉，被学弟吊着打呀qwq

我们连出一条边全为\(2^0\)到\(2^(r - 1)\)的链出来，\(2^r\)是L的最高位

然后我们通过增加一些点到N的边权X，使得数量为L且合法即可。。

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
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 - '0' + c;
        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 L,N;
void add(int u,int v,int c) {
    out(u);space;out(v);space;out(c);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
    read(L);
    int r;
    for(r = 20 ; r >= 0 ; --r) {
        if(L & (1 << r)) break;
    }
    N = r + 1;out(N);space;int M = r * 2 + __builtin_popcount(L) - 1;out(M);enter;
    for(int i = 1 ; i <= N - 1 ; ++i) {
        add(i,i + 1,(1 << i - 1));add(i,i + 1,0);
    }
    for(int i = 1 ; i <= N - 1 ; ++i) {
        if(L & (1 << i - 1)) {
            L ^= (1 << i - 1);
            add(i,N,L);
        }
    }
}

E - Stop. Otherwise...

题解

计数题

加和的限制相当于有几个数两个中只能出现一个

算一个dp[i][j]表示把i个数分成j段，每一段两种染色方式染色的方案数

然后枚举有多少数用来给两种方式染色，有多少数给怎么放都没事的数

如果有出现两次就不合法的数，那么就先算不存在它的排列，再算存在它一个的排列

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define MAXN 1000005
#define mo 999999137
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
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) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int fac[4005],inv[4005],invfac[4005],N,K,vis[4005],pw[4005];
int dp[4005][4005];
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int C(int n,int m) {
    if(n < m) return 0;
    if(n < 0 || m < 0) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    inv[1] = 1;pw[0] = 1;
    for(int i = 2 ; i <= 4000 ; ++i) {
        inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    }
    fac[0] = invfac[0] = 1;
    for(int i = 1 ; i <= 4000 ; ++i) {
        fac[i] = mul(fac[i - 1],i);invfac[i] = mul(invfac[i - 1],inv[i]);
        pw[i] = mul(pw[i - 1],2);
    }
    read(K);read(N);
    dp[0][1] = 1;
    for(int i = 1 ; i <= N ; ++i) dp[i][1] = 2;
    dp[0][0] = 1;
    for(int j = 2 ; j <= K ; ++j) {
        int s = 1;
        dp[0][j] = 1;
        for(int i = 1 ; i <= N ; ++i) {
            dp[i][j] = inc(mul(s,2),dp[i][j - 1]);
            s = inc(s,dp[i][j - 1]);
        }
    }
    for(int i = 2 ; i <= 2 * K ; ++i) {
        int cnt = 0,t = 0,rest = 0;
        memset(vis,0,sizeof(vis));
        for(int j = 1 ; j <= K ; ++j) {
            if(!vis[j] && j != i - j && i - j >= 1 && i - j <= K) {vis[j] = vis[i - j] = 1;++cnt;}
        }
        if(i % 2 == 0) ++t;
        rest = K - cnt * 2 - t;
        int ans = 0;
        for(int j = N ; j >= 0 ; --j) {
            ans = inc(ans,mul(dp[j][cnt],rest ? C(N - j - 1 + rest,rest - 1) : (N - j == 0)));
        }
        if(t) {
            for(int j = N - 1; j >= 0 ; --j) {
                ans = inc(ans,mul(dp[j][cnt],rest ? C(N - 2 - j + rest,rest - 1) : (N - 1 - j == 0)));
            }
        }
        out(ans);enter;
    }
    return 0;
}

F - Revenge of BBuBBBlesort!

题解

我们按照逆操作考虑，容易发现是1-N顺序排列

每次交换\(a_{i - 1} < a_{i} < a_{i + 1}\)使得它变成排列p

这样原始排列每次交换的时候相邻两个不会都进行操作，所以排列p如果合法的话一定是

p[i] != i p[i + 1] == i + 1 p[i + 2] != i + 2....这样的排列

所以我们只要对p排列，顺序找出区间l,r保证l + 1,l + 3...r - 1是固定不动的

然后找出剩下往左往右的点，保证往左的点是顺序的，往右的点是顺序的即可

代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define enter putchar('\n')
#define space putchar(' ')
#define fi first
#define se second
#define MAXN 300005
//#define ivorysi
#define pii pair<int,int>
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) {putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int N;
int p[MAXN],v[MAXN],cnt,id[MAXN],dir[MAXN];
bool vis[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(p[i]);
    int la = 1;
    while(1) {
        while(p[la] == la && la <= N) ++la;
        if(la > N) break;
        int ed = la;
        while(1) {
            if(ed <= N && ((ed - la) & 1) && p[ed] == ed) ++ed;
            else if(ed <= N && !((ed - la) & 1) && p[ed] != ed) ++ed;
            else break;
        }
        --ed;
        if(p[ed] == ed) --ed;
        if(ed == la) {puts("No");return;}
        cnt = 0;
        for(int i = la ; i <= ed ; ++i) v[++cnt] = p[i];
        sort(v + 1,v + cnt + 1);
        for(int i = 1 ; i <= cnt ; ++i) {
            if(v[i] != i + la - 1) {puts("No");return;}
        }
        cnt = 0;
        for(int i = la ; i <= ed ; i += 2) {
            v[++cnt] = p[i];
            id[p[i]] = cnt;
        }
        sort(v + 1,v + cnt + 1);
        for(int i = la ; i <= ed ; i += 2) {
            int t = lower_bound(v + 1,v + cnt + 1,p[i]) - v;
            if(t <= id[p[i]]) dir[i] = 0;
            else dir[i] = 1;
        }
        int L = 0,R = 0;
        for(int i = la ; i <= ed ; i += 2) {
            if(dir[i] == 0) {
                if(p[i] < L) {puts("No");return;}
                L = p[i];
            }
            else {
                if(p[i] < R) {puts("No");return;}
                R = p[i];
            }
        }
        la = ed + 1;
    }
    puts("Yes");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}