牛客OI周赛14 普及组

Prologue

菜的真实，普及都 AK 不掉..

---

A String

看来 PJ T1 考字符串读入成铁上钉钉了？

考虑开桶 \(a\) ，记录 ASCII 为 \(i\) 的字符是否出现即可。

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int maxn = 100000 + 7;
const int maxm = 200000 + 7;

int n, m;

#ifdef Graph_Define
int Head[maxn], to[maxm], Next[maxm], tot;

void addedge(int x, int y) {
    to[++tot] = y, Next[tot] = Head[x], Head[x] = tot;
}
void add(int x, int y) {
    addedge(x, y);
    addedge(y, x);
}
#endif

template < typename Tp >
void read(Tp &x) {
    x = 0;char ch = 1;int fh = 1;
    while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if(ch == '-') fh = -1, ch=getchar();
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    x *= fh;
}

string s;
bool a[200];
void Init(void) {
    cin >> s;
}

int ans;

void Work(void) {
    for(int i = 0; i < (int)s.size(); i++) {
        if(!a[s[i]]) ++ans;
        a[s[i]] = 1;
    }
    cout << ans << '\n';
}

int main() {
    Init();
    Work();
    return 0;
}

---

B Number

发现 \(a_i \le 10^9\) ，考虑到 \(2^{22} >> 10^9\) ，所以对这些数暴力分解，枚举到其 \(22\) 次方的和即可。

#include<bits/stdc++.h>
using namespace std;

#define int long long

typedef long long LL;

const int maxn = 100000 + 7;
const int maxm = 200000 + 7;

int n, m;

#ifdef Graph_Define
int Head[maxn], to[maxm], Next[maxm], tot;

void addedge(int x, int y) {
    to[++tot] = y, Next[tot] = Head[x], Head[x] = tot;
}
void add(int x, int y) {
    addedge(x, y);
    addedge(y, x);
}
#endif

template < typename Tp >
void read(Tp &x) {
    x = 0;char ch = 1;int fh = 1;
    while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if(ch == '-') fh = -1, ch=getchar();
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    x *= fh;
}

int T;

void Init(void) {
    read(T);
}

int ans;

int fpow(int x, int p) {
    int res = 1;
    while(p) {
        if(p & 1) res = res * x;
        p >>= 1, x = x * x;
    }
    return res;
}

bool check(int x) {
    vector < int > v;
    int copy = x;
    while(copy) {
        v.push_back(copy % 10);
        copy /= 10;
    }
    int k = (int)v.size();
    int times = 0;
    while(copy < x && times <= 22) {
        copy = 0;
        for(auto i : v) {
            copy += fpow(i,times);
            if(copy > x) return false;
        }
        if(copy == x) return true;
        if(copy > x) return false;
        ++times;
    }
    return false;
}

void Work(void) {
    while(T--) {
        read(n);
        if(check(n)) ++ans;
    }
    printf("%lld\n", ans);
}

signed main() {
    Init();
    Work();
    return 0;
}

---

C Tree

换根树形 DP。

先钦定 \(1\) 为根，把 dep 都跑出来。

假设当前 \(x\) 为根，向 \(y\) 换根时，有 \(n - size_y\) 的点距离增加一，有 \(size_y\) 个点距离减少一。

所以搜一遍就出来了。

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int maxn = 1000000 + 7;
const int maxm = 2000000 + 7;
const int INF = 0x3f3f3f3f;

int n, m;

#define Graph_Define
#ifdef Graph_Define
int Head[maxn], to[maxm], Next[maxm], tot;

void addedge(int x, int y) {
    to[++tot] = y, Next[tot] = Head[x], Head[x] = tot;
}
void add(int x, int y) {
    addedge(x, y);
    addedge(y, x);
}
#endif

template < typename Tp >
void read(Tp &x) {
    x = 0;char ch = 1;int fh = 1;
    while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if(ch == '-') fh = -1, ch=getchar();
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    x *= fh;
}

void Init(void) {
    read(n);
    for(int i = 1, x, y; i < n; i++) {
        read(x); read(y);
        add(x, y);
    }
}

int dep[maxn], size[maxn], sum[maxn];

void dfs1(int x, int f, int dp) {
    size[x] = 1, dep[x] = dp, sum[x] += dp;
    for(int i = Head[x]; i; i = Next[i]) {
        int y = to[i];
        if(y == f) continue;
        dfs1(y, x, dp + 1);
        size[x] += size[y], sum[x] += sum[y];
    }
}

int ans = INF;

void dfs2(int x, int f, int val) {
    ans = min(ans, val);
    for(int i = Head[x]; i; i = Next[i]) {
        int y = to[i];
        if(y == f) continue;
        dfs2(y, x, val + n - size[y]  - size[y]);
    }
}

void Work(void) {
    dfs1(1, 0, 0);
    dfs2(1, 0, sum[1]);
    printf("%d\n", ans);
}

int main() {
    Init();
    Work();
    return 0;
}