atcoder beginner contest 234题解

最新推荐文章于 2023-07-20 21:08:26 发布

原创最新推荐文章于 2023-07-20 21:08:26 发布 · 558 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#动态规划 #c++ #算法

abc&cf 专栏收录该内容

3 篇文章

订阅专栏

本文提供了AtCoder编程比赛ABC234中六个问题的解决方案，涵盖简单计算、几何距离计算、位运算、堆应用、等差数列查找及组合数学+动态规划等多种算法题型。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

atcoder abc234 a-f

A.Weired Function
简单的计算，记得开long long

#include<iostream>
#include<cmath>
#define int long long
using namespace std;
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int x;
    cin>>x;
    int x1 = pow(x,2) + 2 * x + 3;
    //cout<<x1<<"|||\n";
    int x2 = x1 + x;
    //cout<<x2<<"|||\n";
    int x3 = pow(x2,2) + 2 * x2 + 3;
    //cout<<x3<<"|||\n";
    int x4 = pow(x1,2) + 2 * x1 + 3;
    //cout<<x4<<"|||\n";
    int x5 = x3 + x4;
    //cout<<x5<<"|||\n";
    int ans = pow(x5,2) + 2 * x5 + 3;
    cout<<ans;
}

B.Longest Segment
$O(n^2)$ 寻找最大直线距离

#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
vector<pair<int,int>> v;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++ ){
        int x,y;
        scanf("%d %d",&x,&y);
        v.push_back({x,y});
    }
    double maxx = 0;
    for(int i = 0;i < v.size();i ++ ){
        for(int j = 0;j < v.size();j ++ ){
            maxx = max(maxx,sqrt(abs(v[i].second - v[j].second) * abs(v[i].second - v[j].second) + abs(v[i].first - v[j].first) * abs(v[i].first - v[j].first)));
    }
    }
    printf("%.10lf",maxx);
}

C.Happy New Year
寻找一串由0和2构成的数字中第n个数
本来是在oeis上找到的规律，但是，t了
后来发现位运算就可以解决

#include<iostream>
#include <cmath>
#include<cstring>
#include<algorithm>
#define int long long
using namespace std;
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int x;
    cin>>x;
    string ans;
    while(x){
        if(x & 1) ans += '2';
        else ans += '0';
        x >>= 1;
    }
    reverse(ans.begin(),ans.end());
    cout<<ans;
}

D - Prefix K-th Max
寻找一串数字中第n大的数
大根堆的运用

#include<iostream>
#include <cmath>
#include<cstring>
#include<algorithm>
#include<queue>
//#define int long long
using namespace std;
const int N = 5e5 + 10;
int a[N];
priority_queue<int,vector<int>,greater<int>> q;
int  main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    for(int i = 1;i <= n;i ++ ){
        //int x;
        cin>>a[i];
        if(i <= k){
            q.push(a[i]);
        }
    }
    cout<<q.top()<<"\n";
    for(int i = k + 1;i <= n;i ++ ){
        if(a[i] > q.top()){
            q.pop();
            q.push(a[i]);
        }
        if(q.size() >= k) cout<<q.top()<<"\n";
    }
}

E - Arithmetic Number
寻找最小的大于n的数，并且该数的各个数位组成的数组是等差数列
从-9到9枚举公差，将最小的符合条件的数输出

#include<iostream>
#include <cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#define int long long
using namespace std;
const int N = 20;
int a[N];
int b[N];
int cnt,ans;
int n;
bool check(int n){
    int d = a[2] - a[1];
    for(int i = 1;i < n;i ++ ){
        if(a[i + 1] - a[i] != d) return 0;
    }
    return 1;
}
bool check1(int x){
    int cnt = 0;
    memset(b,0,sizeof b);
    while(x){
        b[++ cnt] = x % 10;
        x /= 10;
    }
    int d = b[2] - b[1];
    for(int i = 1;i < cnt;i ++ ){
        if(b[i + 1] - b[i] != d) return 0;
    }
    return 1;
}
void check(int x,int d){
    //cout<<x<<"|||\n";
    int t = x;
    for(int i = 2;i <= cnt;i ++ ){
        x -= d;
        if(x < 0 || x > 9) return;
        t = t * 10 + x;
    }
    if(check1(t) && t >= n){
        //cout<<t<<"|||\n";
        ans = min(ans,t);
        //cout<<ans<<"|||\n";
    }
}
//int b[10005];
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    //int x;
    cin>>n;
    int te = n;
    while(n){
        a[++ cnt] = n % 10;
        n /= 10;
    }
    n = te;
    reverse(a + 1,a + cnt + 1);
    if(check(n)) cout<<n;
    else{
        ans = 1e18;
        for(int i = -9;i <= 9;i ++ ){
            check(a[1],i);
            check(a[1] + 1,i);
        }
        cout<<ans<<"\n";
    }
    
}

F - Reordering
给出一串字符，求解该字符串的子集个数
组合数学+dp
第一维表示已经用了多少个字母表中的字母
第二维表示用了多少个字母
k是当前第i个字母表中的字母在串中出现的次数
那么可以推断 $f[i + 1][j] = f[i + 1][j] + f[i][j - k] * C_{j}^{k}$

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define int long long
using namespace std;
const int mod = 998244353;
const int MX = 5005;
int f[30][MX];
int cnt[30];
int fac[MX];
int inv[MX];
int finv[MX];
void fac_init(){
    fac[0] = fac[1] = 1;
    inv[1] = 1;
    finv[0] = finv[1] = 1;
    for(int i = 2;i < MX;i ++ ){
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = mod - mod / i * inv[mod % i] % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}
int c(int x,int y){
    if(x < y || x < 0 || y < 0) return 0;
    return fac[x] * finv[y] % mod * finv[x - y] % mod;
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    fac_init();
    string str;
    cin>>str;
    //cout<<str;
    int t = str.size();
    for(int i = 0;i < t;i ++ ){
        cnt[str[i] - 'a'] ++;
    }
    f[0][0] = 1;
    for(int i = 0;i < 26;i ++ ){
        for(int j = 0;j <= t;j ++ ){
            for(int k = 0;k <= min(j,cnt[i]);k ++ ){
                f[i + 1][j] += f[i][j - k] * c(j,k);
                f[i + 1][j] %= mod;
            }
        }
    }
    int ans = 0;
    for(int i = 1;i <= t;i ++ ){
        ans += f[26][i];
        ans %= mod;
    }
    cout<<ans;
}