Codeforces 136A-Presents(题意)

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A. Presents
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited n his friends there.

If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to n. Petya remembered that a friend number i gave a gift to a friend number pi. He also remembered that each of his friends received exactly one gift.

Now Petya wants to know for each friend i the number of a friend who has given him a gift.

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the quantity of friends Petya invited to the party. The second line contains nspace-separated integers: the i-th number is pi — the number of a friend who gave a gift to friend number i. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.

Output

Print n space-separated integers: the i-th number should equal the number of the friend who gave a gift to friend number i.

Sample test(s)
input
4
2 3 4 1
output
4 1 2 3
input
3
1 3 2
output
1 3 2
input
2
1 2
output
1 2
题意很啰嗦,但很简单,现在给一个数组a[]有关系 ans[a[i]]=i; 输出ans[] 即可。
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
#define LL long long
int a[110],ans[110];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=1;i<=n;i++)
		{
			scanf("%d",a+i);
			ans[a[i]]=i;
		}
		for(int i=1;i<=n;i++)
			if(i!=n)
			printf("%d ",ans[i]);
		    else
			printf("%d\n",ans[i]);
	}
	return 0;
}
### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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