Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B: Conformity

Frosh commencing their studies at Waterloo have diverse interests, as evidenced by their desire to take various combinations of courses from among those available.

University administrators are uncomfortable with this situation, and therefore wish to offer a conformity prize to frosh who choose the most popular combination of courses. How many frosh will win the prize?

The input consists of several test cases followed by a line containing 0. Each test case begins with an integer 1 ≤ n ≤ 10000, the number of frosh. For each frosh, a line follows containing the course numbers of five distinct courses selected by the frosh. Each course number is an integer between 100 and 499.

The popularity of a combination is the number of frosh selecting exactly the same combination of courses. A combination of courses is considered most popular if no other combination has higher popularity. For each line of input, you should output a single line giving the total number of students taking some combination of courses that is most popular.

Sample Input

3
100 101 102 103 488
100 200 300 101 102
103 102 101 488 100
3
200 202 204 206 208
123 234 345 456 321
100 200 300 400 444
0

Output for Sample Input

2
3

---

#include <iostream>

#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;

int main() {
	int n, A[5];
	char buf[20];
	while(scanf("%d", &n) == 1 && n) {
		string str;
		map<string, int> record;
		while(n--) {
			scanf("%d %d %d %d %d", &A[0], &A[1], &A[2], &A[3], &A[4]);
			sort(A, A+5);
			sprintf(buf, "%d%d%d%d%d", A[0], A[1], A[2], A[3], A[4]);
			str = buf;
			record[str]++;
		}
		int max = 0, maxNum = 0;
		for(map<string, int>::iterator i = record.begin(); i != record.end(); i++) {
			if(i->second > max)
				max = i->second, maxNum = 0;
			if(i->second == max)
				maxNum += max;
		}
		printf("%d\n", maxNum);
	}
    return 0;
}