Namomo Winter Camp 2023 Day 2

A - Mix Juice

排个序再求和就好

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

int read(){...}

int32_t main(){
    int n = read() , k = read();
    vector<int> a(n);
    for( auto & i : a ) i = read();
    sort( a.begin() , a.end() );
    int res = 0;
    for( int i = 0 ; i < k ; i ++ ) res += a[i];
    cout << res << "\n";

    return 0;
}

B - One Quadrillion and One Dalmatians

进制转换，因为没有0所以要特殊处理一下

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

int32_t main(){
    int n; cin >> n;
    vector<int>res;
    do{
        n-- , res.push_back( n%26 ) , n /= 26;
    }while( n > 0 );
    std::reverse(res.begin(), res.end());
    for( int i : res )
        cout << char(i+'a');
}

C - Replacing

#include<bits/stdc++.h>
#define int long long
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;
}

const int N = 1e5+5;
int v[N] , res = 0;

int32_t main(){
    for( int n = read() , x ; n ; n -- ) x = read() , res += x , v[x] ++;
    for( int q = read() , b , c ; q ; q -- )
        b = read() , c = read() , res += (c-b) * v[b] , v[c] += v[b] , v[b] = 0 , printf("%lld\n" , res );
    return 0;
}

D - Red Scarf

因为保证了偶数，所以把所有的数异或起来就等于原数每个数异或起来。然后再异或剩余的数就是原数

#include<bits/stdc++.h>
#define int long long
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;
}

int32_t main(){
    int n = read() , sum = 0;
    vector<int> a(n);
    for( auto & i : a ) i = read() , sum ^= i;
    for( auto i : a ) printf("%lld " , sum ^ i );
    return 0;
}

F - Minor Change

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

int32_t main(){
    ios::sync_with_stdio(false) , cin.tie();
    string s , t;
    int res = 0;
    cin >> s >> t;
    for( int i = 0 ; i < s.size() ; i ++ )
        if( s[i] != t[i] ) res ++;
    cout << res << "\n";
    return 0;
}

G - Tsundoku

排序，求前缀和，枚举一个买多少二分另一个能买多少，然后判断一下就好

#include<bits/stdc++.h>
#define int long long
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;
}

int32_t main(){
    int n = read() , m = read() , k = read() , res = 0;
    vector<int> a(n+1) , b(m+1);
    for( int i = 1 ; i <= n ; i ++ ) a[i] = read();
    for( int i = 1 ; i <= m ; i ++ ) b[i] = read();
    for( int i = 1 ; i <= n ; i ++ ) a[i] += a[i-1];
    for( int i = 1 ; i <= m ; i ++ ) b[i] += b[i-1];
    for( int i = 0 , t , j ; i <= n ; i ++ ){
        if( a[i] > k ) break;
        t = k - a[i];
        j = upper_bound( b.begin() , b.end() , t ) - b.begin() - 1;
        res = max( res , i + j );
    }
    cout << res << "\n";
}

H - Sum of Divisors

做一个类似埃筛的东西，去标记一个数的倍数，这样可以\(O(n\log n)\)的求出所有数有多少个因子，然后再\(O(n)\)的算一边就好

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


int32_t main(){
    int n;
    cin >> n;
    vector<int> f(n+1);
    for( int i = 1 ; i <= n ; i ++ )
        for( int j = i ; j <= n ; j += i ) f[j]++;
    int res = 0;
    for( int i = 1 ; i <= n ; i ++ ) res += i * f[i];
    cout << res << "\n";
    return 0;
}

I - NEQ(补题)

首先这题赛时没出就是自己傻逼了，预处理阶乘的是否忘记取模

首先我们不考虑\(B_i\)，只考虑\(A_i\)的情况，显然有\(A_m^n\)中情况，然后我们要计算出对于每一个序列\(A\)有多少个序列\(B\)，易知对于每种序列\(A\)符合条件的序列\(B\)的数量是相等的。

直接计算合法的序列\(B\)比较困难，我们用序列\(B\)的总数减去不合法的数量就是合法的数量。

用到了一个组合容斥，暂时还不太会，后面补一下。

\[cnt=\sum_{k=0}^n (-1)^k\times C(n,k) \times A(m-k,n-k) \]

最后的答案就是\(cnt\times A(m,n)\)

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

const int N = 5e5+5 , mod = 1e9+7;
int fact[N] , invFact[N] , n , m , res = 0;

int power(int x,int y)
{
    int ans =1;
    while(y)
    {
        if( y & 1 ) ans = ans * x % mod;
        x= x * x %mod;
        y>>=1;
    }
    return ans;
}

int inv( int x ){
    return power( x , mod - 2 );
}

int A( int x , int y ){
    return fact[x] * invFact[x-y] % mod;
}

int C( int x , int y ){
    return fact[x] * invFact[x-y] % mod * invFact[y] % mod;
}

int32_t main(){
    fact[0] = 1 , invFact[0] = inv(1);
    for( int i = 1 ; i < N ; i ++ ) fact[i] = fact[i-1] * i % mod , invFact[i] = inv(fact[i]);
    cin >> n >> m;
    for(int i = 0;i <= n;i++){
        if(i&1) res=(res-C(n,i)*A(m-i,n-i)%mod+mod)%mod;
        else res=(res+C(n,i)*A(m-i,n-i)%mod)%mod;
    }
    cout << res * A(m,n) % mod;
}