Codeforces Round #847 (Div. 3)

1|0A. Polycarp and the Day of Pi

先纯一下圆周率前30位，然后暴力就好了

#include <bits/stdc++.h>

using std::cin;
using std::cout;
using std::string;

int read(){...}

const string tmp = "314159265358979323846264338327";

void solve(){
    string s; cin >> s;
    int res = 0;
    for( int i = 0 ; i < std::min( s.size() , tmp.size() ) ; i ++ ){
        if( s[i] == tmp[i] ) res ++;
        else break;
    }
    printf("%d\n" , res );

}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

2|0B. Taisia and Dice

令s-r为第一个数，保证序列不上升，且和为s。dfs枚举一下就行。

#include <bits/stdc++.h>

using std::cin;
using std::cout;
using std::string;

const int N = 55;
int a[N] , n , s , r ;
bool flag ;

int read(){...}

void dfs( int i , int sum ){
    if( flag ) return;
    if( i == n + 1 ){
        if( sum != 0 ) return ;
        flag = true;
        for( int j = 1 ; j <= n ; j ++ ) printf("%d%c" , a[j] , " \n"[j==n] );
        return;
    }
    for( int j = std::min( a[i-1] , sum ) ; j >= 1  ; j -- )
        a[i] = j , dfs( i+1 , sum - j );
    return ;
}

void solve(){
    n = read() , s = read() , r = read() , flag = false;
    memset( a , 0 , sizeof a ) , a[1] = s - r;
    dfs( 2 , r );
    return ;
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

3|0C. Premutation

每个位置的数一定出现了n-1次，而没有出现该数的序列就是这个数缺失的序列，这个序列后面的数向前错一位保证对其

#include <bits/stdc++.h>

using namespace std;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

void solve(){
    int n = read();
    vector< vector<int> > a( n ,  vector<int>(n-1) );
    for( auto & it : a )
        for( auto &i : it ) i = read();
    vector<int> vis(n);
    for( int i = 0 ; i < n ; i ++ ){
        map<int,int> st;
        for( int j = 0 ; j < n ; j ++ ){
            if( vis[j] ) st[ a[j][i-1] ] ++;
            else st[ a[j][i] ] ++;
        }
        for( auto [ k , v] : st ){
            if( v == 1 ){
                for( int j = 0 ; j < n ; j ++ )
                    if( vis[j] == 0 && a[j][i] == k ) vis[j] = 1;
            }
            else printf("%d " , k );
        }
    }
    printf("\n");
    return;
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

但是看了jiangly发现有更简单的做法，当找到出第一个数，第一数确实的序列是[2,n]的完整序列

#include <bits/stdc++.h>

using namespace std;


int read() {...}

void solve() {
    int n = read() , t;
    vector<vector<int> > a(n, vector<int>(n - 1));
    for (auto &it: a)
        for (auto &i: it) i = read();
    map<int, int> st;
    for (auto it: a) st[it[0]]++;
    for (auto [k, v]: st) if (v == n - 1) t = k;
    for (auto it: a) {
        if (it[0] == t) continue;
        printf("%d ", t);
        for (auto i: it) printf("%d ", i);
        printf("\n");
        return;
    }
    return;
}

int32_t main() {
    for (int t = read(); t; t--)
        solve();
    return 0;
}

4|0D. Matryoshkas

数字很少，直接暴力就好。

#include <bits/stdc++.h>

using namespace std;

int read(){...}

void solve(){
    int n = read() , res = 0 , t;
    map<int,int> st;
    for( int i = 1 , x ; i <= n ; i ++ ) x = read() , st[x]++;
    while( !st.empty() ){
        vector<int> erase;
        res ++ , t = -1;
        for( auto &[k,v] : st ){
            if( k == t+1 || t == -1 ) t = k , v --;
            else break;
            
            if( v == 0 ) erase.push_back(k);
        }
        for( auto it : erase ) st.erase(it);
    }
    printf("%d\n" , res );
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

5|0E. Vlad and a Pair of Numbers

我这里是先用lowbit把原来的数拆开，然后这些数保证了异或为x，然后因为 $x=\frac{a+b}{2}$ ，所以 $a+b-x=x$ ，也就是说还有x保证和是a+b，这部分因为异或是0所以要平均分，即 $\frac{x}{2}$ 并且些数lowbit拆开后不能与x拆开的数重复。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

#define lowbit(x) ( x &-x )

void solve(){
    int x = read();
    if( x & 1 ) {
        printf("-1\n");
        return ;
    }
    set<int> a , b ;
    for( int t = x , p ; t > 0 ; ) p = lowbit(t) , a.insert(p) , t -= p;
    for( int t = (x >> 1) , p ; t > 0 ; ) p = lowbit(t) , b.insert(p) , t -= p;
    for( auto i : b ){
        if( a.count(i) ) {
            printf("-1\n");
            return ;
        }
    }
    int A = 0 , B = 0;
    for( auto i : a ) A += i ;
    for( auto i : b ) A += i , B += i ;
    cout << A << " " << B << "\n";
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

赛后发现有 $O(1)$ 的做法，观察我构造的值， $a=\frac{x}{2},b=\frac{3x}{2}$ 直接验证就好

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read(){...}

void solve(){
    int x = read() , a = x / 2 , b = a * 3 ;
    if(( a ^ b) != x || a + b != x * 2 ) printf("-1\n");
    else printf("%d %d\n" , a , b );
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

6|0F. Timofey and Black-White Tree(补题)

直接bfs，在保证最优性的情况下，层数下降的非常快，大概是 $n\times(\frac11+\frac12+\frac12+\frac14+\frac14+\frac14+\frac14+\frac18 \cdots)$ 的情况。知道了这个性质我就想写了一个dfs

#include <bits/stdc++.h>

using namespace std;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

int n , ans;
vector<int> c , res , color;
vector<vector<int>> e;

void dfs( int u , int fa , int dis ){
    if( dis >= ans ) return;
    for( auto v : e[u] ){
        if( v == fa ) continue;
        if( color[v] == 1 ) {
            ans = min( ans , dis + 1 );
            return;
        }
        dfs( v , u , dis + 1 );
    }
}

void solve(){
    n = read() , ans = INT_MAX;
    c = vector<int>(n) , res.clear() , color = vector<int>(n+1 , 0 ) , e = vector<vector<int>>( n+1 );
    for( auto &i : c ) i = read();
    for( int i = 1 , u , v ; i < n ; i ++ ){
        u = read() , v = read();
        e[u].push_back(v) , e[v].push_back(u);
    }
    color[ c[0] ] = 1;
    for( int i = 1 ; i < n ; i ++ ){
        color[ c[i] ] = 1;
        dfs( c[i] , -1 , 0 );
        cout << ans << " ";
    }
    cout << "\n";
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

然后TLE20，觉得可能dfs的问题又写了一下bfs

#include <bits/stdc++.h>

using namespace std;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

int n , ans;
vector<int> c , res , color;
vector<vector<int>> e;

#define mp make_pair

void solve(){
    n = read() , ans = INT_MAX;
    vector<int> c(n) , color( n+1 , 0 );
    vector< vector<int> > e(n+1);
    for( auto &i : c ) i = read();
    for( int i = 1 , u , v ; i < n ; i ++ ){
        u = read() , v = read();
        e[u].push_back(v) , e[v].push_back(u);
    }
    color[ c[0] ] = 1;

    for( int i = 1 ; i < n ; i ++ ){
        color[ c[i] ] = 1;
        unordered_set<int> vis;
        queue< pair<int,int> > q;
        q.push( mp( c[i] , 0 ) ) , vis.insert(c[i]);
        while( !q.empty() ){
            auto [ u , dis ] = q.front(); q.pop();
            if( dis >= ans ) break;
            bool flag = false;
            for( auto v : e[u] ){
                if( vis.count(v) ) continue;
                if( color[v] == 1 ) {
                    ans = min( ans , dis + 1 );
                    flag = true;
                    break;
                }
                q.emplace( v , dis + 1 ) , vis.emplace(v);
            }
            if( flag ) break;
        }
        printf("%d " , ans );
    }
    printf("\n");
    return;
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

还是TLE20，我们换个角度来思考一下，只有新增加的点可能会影响答案，所以我们维护一个每个点到黑点的最短距离，没次加入一个点后，用bfs去更像一下整张图的最短距离

#include <bits/stdc++.h>

using namespace std;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

int n , ans;
vector<int> c , res , color;
vector<vector<int>> e;

#define mp make_pair

void solve(){
    n = read() , ans = INT_MAX;
    vector<int> c(n) , color( n+1 , 0 );
    vector< vector<int> > e(n+1);
    for( auto &i : c ) i = read();
    for( int i = 1 , u , v ; i < n ; i ++ ){
        u = read() , v = read();
        e[u].push_back(v) , e[v].push_back(u);
    }
    vector<int> dis( n+1 , n );
    for( auto i : c ){
        ans = min( ans , dis[i] );
        if( ans < n ) printf( "%d " , ans );
        queue<int> q;
        q.push( i ) , dis[i] = 0;
        while( !q.empty() ){
            auto u = q.front() ; q.pop();
            if( dis[u] + 1 >= ans ) break;
            for( auto v : e[u] ){
                if( dis[u] + 1 >= dis[v] ) continue;
                dis[v] = dis[u] + 1 , q.push( v );
            }
        }
    }
    printf("\n");
    return;
}

int32_t main(){
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}