【AtCoder】AGC005

其他 2019-05-19 19:58:02 阅读次数: 0

AGC005

A - STring

用一个栈，如果遇到S就弹入，如果遇到T栈里有S就弹出栈顶，否则T在最后的串里，最后计算出的T和栈里剩的S就是答案

#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 eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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);
}
char s[MAXN];
int top,N;
void Solve() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
        if(s[i] == 'T') {
            if(top) --top;
            else ++ans;
        }
        else {
            ++top;
        }
    }
    ans += top;
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

B - Minimum Sum

就是单调栈求出左边最靠近它且比它小的数，右边最靠近且比它小的数，就可以计算出以这个点为最小值的区间有多少了

#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 eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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,a[MAXN],l[MAXN],r[MAXN];
int sta[MAXN],top;
int64 ans = 0;
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    for(int i = 1 ; i <= N ; ++i) {
        while(top && a[sta[top]] > a[i]) --top;
        l[i] = sta[top] + 1;
        sta[++top] = i;
    }
    top = 0;sta[0] = N + 1;
    for(int i = N ; i >= 1 ; --i) {
        while(top && a[sta[top]] > a[i]) --top;
        r[i] = sta[top] - 1;
        sta[++top] = i;
    }
    for(int i = 1 ; i <= N ; ++i) {
        ans += 1LL * (i - l[i] + 1) * (r[i] - i + 1) * a[i];
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

C - Tree Restoring

最大的A肯定是直径，如果直径是奇数，看看最小值是不是两个，其余的长度必须都大于等于2个

如果直径是偶数，看看最小值是不是1个，其余的长度必须大于等于两个

#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 eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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 D,N,a[MAXN],cnt[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);cnt[a[i]]++;}
    for(int i = 1 ; i <= N ; ++i) D = max(D,a[i]);
    if(D & 1) {
        if(cnt[D / 2 + 1] != 2) {puts("Impossible");return;}
        int res = 0;
        for(int i = D / 2 + 1 ; i <= D ; ++i) {
            res += cnt[i];
            if(cnt[i] < 2) {puts("Impossible");return;}
        }
        if(res != N) {puts("Impossible");return;}
    }
    else {
        if(cnt[D / 2] != 1) {puts("Impossible");return;}
        int res = 1;
        for(int i = D / 2 + 1 ; i <= D ; ++i) {
            res += cnt[i];
            if(cnt[i] < 2) {puts("Impossible");return;}
        }
        if(res != N) {puts("Impossible");return;}
    }
    puts("Possible");return;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - ~K Perm Counting

用经典容斥，设\(g(i)\)为至少i个位置不合法的情况

答案就是\(\sum_{i = 0}^{n}(-1)^{i}g(i)\)

以\([1,2K - 1]\)开头，每\(2K\)个选择一个构成的这样的序列，这个序列当前的数每个能填什么，之和上一个数选择什么有关，这个可以用一个dp完成，然后可以得到每个序列里至少选了\(i\)个方案数

组合起来可以用NTT优化

#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 eps 1e-10
#define MAXN 2005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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 = 924844033,g = 5,MAXL = (1 << 15);

int N,K;
int a[MAXN],W[MAXL + 5],f[2][MAXN][2],p[MAXL + 5];
int ans[MAXL + 5],fac[MAXN];
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);
}
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;
}
void NTT(int *p,int L,int on) {
    for(int i = 1 , j = L >> 1 ; i < L - 1 ; ++i) {
        if(i < j) swap(p[i],p[j]);
        int k = L >> 1;
        while(j >= k) {
            j -= k;
            k >>= 1;
        }
        j += k;
    }
    for(int h = 2 ; h <= L ; h <<= 1) {
        int wn = W[(MAXL + on * MAXL / h) % MAXL];
        for(int k = 0 ; k < L ; k += h) {
            int w = 1;
            for(int j = k ; j < k + h / 2 ; ++j) {
                int u = p[j],t = mul(p[j + h / 2],w);
                p[j] = inc(u,t);
                p[j + h / 2] = inc(u,MOD - t);
                w = mul(w,wn);
            }
        }
    }
    if(on == -1) {
        int invL = fpow(L,MOD - 2);
        for(int i = 0 ; i < L ; ++i) p[i] = mul(p[i],invL);
    }
}
void Solve() {
    read(N);read(K);
    W[0] = 1;W[1] = fpow(g,(MOD - 1) / MAXL);
    for(int i = 2 ; i < MAXL ; ++i) W[i] = mul(W[i - 1],W[1]);
    int len = 1;
    while(len <= N) len <<= 1;
    ans[0] = 1;
    NTT(ans,len,1);
    for(int i = 1 ; i <= 2 * K ; ++i) {
        memset(f,0,sizeof(f));
        int cur = 0;
        f[cur][0][0] = 1;int cnt = 0;
        for(int j = i ; j <= N ; j += 2 * K) {
            memset(f[cur ^ 1],0,sizeof(f[cur ^ 1]));
            for(int h = 0 ; h <= cnt ; ++h) {
                update(f[cur ^ 1][h][0],inc(f[cur][h][0],f[cur][h][1]));
                if(j - K >= 1) update(f[cur ^ 1][h + 1][0],f[cur][h][0]);
                if(j + K <= N) update(f[cur ^ 1][h + 1][1],inc(f[cur][h][0],f[cur][h][1]));
            }
            ++cnt;
            cur ^= 1;
        }
        memset(p,0,sizeof(p));
        for(int j = 0 ; j <= N ; ++j) {
            p[j] = inc(f[cur][j][0],f[cur][j][1]);
        }
        NTT(p,len,1);
        for(int j = 0 ; j < len ; ++j) {
            ans[j] = mul(ans[j],p[j]);
        }
    }
    NTT(ans,len,-1);
    fac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
    int res = 0;
    for(int i = 0 ; i <= N ; ++i) {
        if(i & 1) update(res,MOD - mul(fac[N - i],ans[i]));
        else update(res,mul(fac[N - i],ans[i]));
    }
    out(res);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Sugigma: The Showdown

观察出-1的条件是先手走到一条红边的某个端点，这个红边的两个端点在蓝边树上的距离大于等于3

之后我们可以进行深搜，如果可以到某个点，先手和这个点的距离小于后手到达这个的距离就可以走，否则不行

如果走到特殊点输出-1，否则就是能走到的距离后手最远的距离乘2

#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 eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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);
}
vector<int> to[2][MAXN];
int fa[MAXN][20],dep[MAXN];
int N,X,Y,ans;
bool flag = 0;
int lca(int u,int v) {
    if(dep[u] < dep[v]) swap(u,v);
    int l = 18;
    while(dep[u] > dep[v]) {
        if(dep[fa[u][l]] >= dep[v]) {
            u = fa[u][l];
        }
        --l;
    }
    if(u == v) return u;
    l = 18;
    while(fa[u][0] != fa[v][0]) {
        if(fa[u][l] != fa[v][l]) {
            u = fa[u][l];
            v = fa[v][l];
        }
        --l;
    }
    return fa[u][0];
}
int dist(int u,int v) {
    return dep[u] + dep[v] - 2 * dep[lca(u,v)];
}
void dfs(int u) {
    for(auto t : to[1][u]) {
        if(t != fa[u][0]) {
            dep[t] = dep[u] + 1;
            fa[t][0] = u;
            dfs(t);
        }
    }
}
void dfs1(int u,int f,int d) {
    if(u == Y) return;
    ans = max(ans,dep[u] - 1);
    for(auto t : to[0][u]) {
        if(t == f) continue;
        if(dist(u,t) > 2) flag = 1;
        if(d + 1 < dep[t] - 1) dfs1(t,u,d + 1);
    }
}
void Solve() {
    read(N);read(X);read(Y);
    int a,b;
    for(int i = 1 ; i < N ; ++i) {
        read(a);read(b);
        to[0][a].pb(b);to[0][b].pb(a);
    }
    for(int i = 1 ; i < N ; ++i) {
        read(a);read(b);
        to[1][a].pb(b);to[1][b].pb(a);
    }
    dep[Y] = 1;
    dfs(Y);
    for(int j = 1 ; j <= 19 ; ++j) {
        for(int i = 1 ; i <= N ; ++i) {
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
        }
    }
    dfs1(X,0,0);
    if(flag) {puts("-1");}
    else {out(2 * ans);enter;}
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - Many Easy Problems

我们把一个点的贡献转化为一条边的贡献，因为边的数量是点的数量-1，最后再加上选点方案数\(\binom{n}{k}\)即可

一条边的贡献是\(\binom{n}{k} - \binom{a}{k} - \binom{n - a}{k}\)就是在n个点里选k个点，去掉不合法的情况也就是k个点都在去掉这条边的两个子树里

然后我们要统计的就是\(\binom{a}{k} + \binom{n - a}{k}\)

这个可以转化成\(ans_{k} = \sum_{i = 1}^{n} b_{i} \binom{i}{k}\)

\(ans_{k} = \frac{1}{k!} \sum_{i = 1}^{n} b_{i} i! \frac{1}{(i - k)!}\)
这个数组是可以卷积的，只要把一个倒过来就行
设\(f(i) = \frac{1}{(n - i)!}\)
\(g(i) = b_{i}i!\)
\(h = g * f\)
\(ans_{k} = h(N + k)\)

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define RG register
#define MAXN 200005
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) {putchar('-');x = -x;}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
const int MOD = 924844033,L = (1 << 19),G = 5;
int N,f[L + 5],fac[MAXN],invfac[MAXN],b[L + 5],W[L + 5];
int mul(int a,int b) {return 1LL * a * b % MOD;}
int inc(int a,int b) {a = a + b;if(a >= MOD) a -= MOD;return a;}
int fpow(int x,int c) {
    int t = x,res = 1;
    while(c) {
    if(c & 1) res = mul(res,t);
    t = mul(t,t);
    c >>= 1;
    }
    return res;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(mul(fac[n],invfac[m]),invfac[n - m]);
}
int dfs(int u,int fa) {
    int siz = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        siz += dfs(v,u);
    }
    }
    if(fa != 0) {
    b[siz]++;
    b[N - siz]++;
    }
    return siz;
}
void NTT(int *a,int LEN,int on) {
    for(int i = 1 , j = LEN / 2 ; i < LEN - 1 ; ++i) {
    if(i < j) swap(a[i],a[j]);
    int k = LEN / 2;
    while(j >= k) {
        j -= k;
        k >>= 1;
    }
    j += k;
    }
    for(int h = 2 ; h <= LEN ; h <<= 1) {
    int wn = W[(L + on * L / h) % L];
    for(int k = 0 ; k < LEN ; k += h) {
        int w = 1;
        for(int j = k ; j < k + h / 2 ; ++j) {
        int64 u = a[j],t = 1LL * a[j + h / 2] * w;
        a[j] = (u + t) % MOD;
        a[j + h / 2] = (u - t + 1LL * MOD * MOD) % MOD;
        w = mul(w,wn);
        }
    }
    }
    if(on == -1) {
    int invL = fpow(LEN,MOD - 2);
    for(int i = 0 ; i < LEN ; ++i) a[i] = mul(a[i],invL);
    }
}
void Solve() {
    read(N);
    int u,v;
    for(int i = 1 ; i < N ; ++i) {
    read(u);read(v);add(u,v);add(v,u);
    }
    dfs(1,0);
    fac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[N] = fpow(fac[N],MOD - 2);
    for(int i = N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    for(int i = 0 ; i <= N ; ++i) f[N - i] = invfac[i];
    for(int i = 1 ; i <= N ; ++i) b[i] = mul(b[i],fac[i]);
    W[0] = 1;W[1] = fpow(G,(MOD - 1) / L);
    for(int i = 2 ; i < L ; ++i) W[i] = mul(W[i - 1],W[1]);
    int t = 1;
    while(t <= 2 * N) t <<= 1;
    NTT(b,t,1);NTT(f,t,1);
    for(int i = 0 ; i < t ; ++i) f[i] = mul(f[i],b[i]);
    NTT(f,t,-1);
    for(int i = 1 ; i <= N ; ++i) {
    int ans = mul(f[i + N],invfac[i]);
    ans = inc(mul(N,C(N,i)),MOD - ans);
    out(ans);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

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