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Codeforces Round 958 (Div.2)

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7月11日 Codeforces 实战记录。（1505 -> 1531）

AC数3/6。
比赛跳转：Codeforces Round 958 (Div. 2)

题解

A. Split the Multiset

即求 (n - 1) / (k - 1) 的上取整。

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

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve(){
    int n = read(), k = read();
    cout << (n - 1 + k - 2) / (k - 1) << '\n'; 
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}

B. Make Majority

考虑结束状态：1 的个数严格大于 0 的个数。
所以只考虑 减少1的个数 严格少于 减少0的个数 的操作。
将整块的 0 全部合并为一个 0，再考虑0，1的个数即可。

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

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve(){
    int n = read();
    string s; cin >> s;
    int cnt0 = 0, cnt1 = 0;
    
    for(int i = 0; i < n;){
        if(s[i] == '1') {
            cnt1 ++;
            i++;
        }
        else{
            cnt0++;
            while(s[i] == '0' && i < n){
                i++;
            }
        }	
    }
    // cout << cnt0 << ' ' << cnt1 << '\n';
    if(cnt1 > cnt0) cout << "Yes" << '\n';
    else cout << "No" << '\n';
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}

C. Increasing Sequence with Fixed OR

构造题，注意：
1.移位时记得用 1ll (犯错的人被卡了18min)
2.对 long long 类型 要用 __builtin_popcountll() 函数 （犯错的人被卡了45min）

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

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve() {
    int n = read();
    int k = __builtin_popcountll(n); 

    if(k == 1){
        cout << 1 << '\n';
        cout << n << '\n';
        return;
    }
    
    cout << k + 1 << '\n';
    for(int i = 60; i >= 0; i--){
        if(n >> i & 1){
            cout << (n ^ (1ll << i)) << ' ';
        }
    }
    cout << n << '\n';
}

signed main() {
    int t = read();
    while (t--) solve();
    return 0;
}

D. The Omnipotent Monster Killer

树形dp，状态转移方程：
$$
dp[u][i] = a[u] * i + \sum_{v\in g[u]}\min_{1\leq j\leq m,\ i!= j} dp[v][j]
$$

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#include<bits/stdc++.h>
#define int unsigned long long
using namespace std;

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

const int N = 3e5 + 10, M = 20;
vector<int>g[N];
int a[N], dp[N][M];

void dfs(int u, int fa){
    for(int i = 1; i < M; i++){
        dp[u][i] = a[u] * i;
    }
    for(auto &v : g[u]){
        if(v == fa) continue;
        dfs(v, u);
        
        int pos = -1, minn = 1e18;
        for(int j = 1; j < M; j++){
            if(dp[v][j] < minn){
                minn = dp[v][j];
                pos = j;
            }
        }
        
        int last = 1e18;
        for(int j = 1; j < M; j++){
            if(j == pos) continue;
            last = min(last, dp[v][j]);
        }
        
        for(int i = 1; i < M; i++){
            if(i == pos) dp[u][i] += last;
            else dp[u][i] += minn;
        }
    
    }
}

void solve(){
    int n = read();
    for(int i = 1; i <= n; i++) a[i] = read(), g[i].clear();
    for(int i = 1; i < n; i++){
        int x = read(), y = read();
        g[x].emplace_back(y);
        g[y].emplace_back(x);
    }
    dfs(1, 0);
    cout << *min_element(dp[1] + 1, dp[1] + M) << '\n';
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}