蓝桥杯官网 试题 PREV-106 历届真题 修改数组【第十届】【省赛】【研究生组】【C++】【C】【Java】【Python】四种解法

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Max = 1000010;

int f[Max];
int N;
int getf(int n)
{
    if(f[n] == 0)
        return n;
    return  f[n] = getf(f[n]+1);
}


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N;
    int d;
    for(int i = 1; i <= N; ++i)
    {
        cin >> d;
        int fd = getf(d);
        cout << fd << " ";
        f[d] = fd;
        f[fd] = fd;  // 这句不能漏掉，否则不能跳过已经使用过的数字
    }

    return 0;
}

#include <stdio.h> 
const int N = 1000010;
int p[1000001]={0};
//查找祖宗节点+路径压缩
int find(int x )
{
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}
int main()
{
    int n;
    int i;
    scanf("%d",&n);
    for( i = 0; i < N; i++)
        p[i] = i;
    for( i = 0; i < n; i++)
    {
        int x;
        scanf("%d",&x);
        x = find(x);
        printf("%d ",x);
        p[x] = x+1;
    }
    return 0;
}

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;

public class Main {
	static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	static int n;
	static int[] a = new int[100010];
	static int[] p = new int[1000010];
	public static void main(String[] args) throws IOException {
		n = Integer.parseInt(bf.readLine());
		String[] msg = bf.readLine().split(" ");
		StringBuilder sb = new StringBuilder();
		
		for(int i = 1; i <= n; i++) 	a[i] = Integer.parseInt(msg[i-1]);
		for(int i = 0; i < 1000010; i++)	p[i] = i;
		
		
		for(int i = 1; i <= n; i++) {
			int pa = find(a[i]);
			sb.append(pa+" ");
			p[pa] = pa+1;
		}
		System.out.println(sb.toString().trim());
	}
	private static int find(int x) {
		if(x!=p[x]) p[x] = find(p[x]);
		return p[x];
	}

}

N=100010
p=[i for i in range(N)]
def find(x):
    if p[x]!=x:
        p[x]=find(p[x])
    return p[x]
n=int(input())
cin=list(map(int,input().split()))
for i in range(n):
    x=find(cin[i])
    if i<n-1:
        print(x,end=' ')
    else:
        print(x)
    p[x]=x+1