PAT甲级：1124 Raffle for Weibo Followers (20分)

题干

John got a full mark on PAT. He was so happy that he decided to hold a raffle（抽奖） for his followers on Weibo -- that is, he would select winners from every N followers who forwarded his post, and give away gifts. Now you are supposed to help him generate the list of winners.

Input Specification:

Each input file contains one test case. For each case, the first line gives three positive integers M (≤ 1000), N and S, being the total number of forwards, the skip number of winners, and the index of the first winner (the indices start from 1). Then M lines follow, each gives the nickname (a nonempty string of no more than 20 characters, with no white space or return) of a follower who has forwarded John's post.

Note: it is possible that someone would forward more than once, but no one can win more than once. Hence if the current candidate of a winner has won before, we must skip him/her and consider the next one.

Output Specification:

For each case, print the list of winners in the same order as in the input, each nickname occupies a line. If there is no winner yet, print Keep going... instead.

Sample Input 1:

9 3 2

Imgonnawin!

PickMe

PickMeMeMeee

LookHere

Imgonnawin!

TryAgainAgain

TryAgainAgain

Imgonnawin!

TryAgainAgain

Sample Output 1:

PickMe

Imgonnawin!

TryAgainAgain

Sample Input 2:

2 3 5

Imgonnawin!

PickMe

Sample Output 2:

Keep going...

思路

需要输出keep going… 的时候，一定是起始点没有在范围里，只需判断一下起始点的位置特判输出它。
接下来就直接按照题意进行编写。比较简单。

code

#include <iostream>

#include <string>

#include <vector>

#include <unordered_map>

using namespace std;

int main(){

	int n_user = 0, step = 0, index = 0;

	scanf("%d%d%d", &n_user, &step, &index);

	vector<string> list(n_user);

	unordered_map<string, bool> dic;

	for(int i = 0; i < n_user; i++){

		list[i].resize(25);

		scanf("%s", &list[i][0]);

	}

	if(index - 1 >= list.size()){

		printf("Keep going...");

		return 0;

	}

	for(int i = index - 1; i < list.size(); ){

		if(!dic[list[i]]){

			dic[list[i]] = true;

			printf("%s\n", list[i].c_str());

			i += step;

		}else i++;

	}

	return 0;

}

巴特西

PAT甲级：1124 Raffle for Weibo Followers (20分)

PAT甲级：1124 Raffle for Weibo Followers (20分)

题干

Input Specification:

Output Specification:

Sample Input 1:

Sample Output 1:

Sample Input 2:

Sample Output 2:

思路

code

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