ZOJ-2132-The MostFrequent Number题解

ZOJ-2132-The MostFrequent Number

---

Time Limit: 5 Seconds      Memory Limit: 1024 KB

---

Seven (actuallysix) problems may be somewhat few for a contest. But I am really unable todevise another problem related to Fantasy Game Series. So I make up an veryeasy problem as the closing problem for this contest.

Given a sequenceof numbers A, for a number X if it has the most instances (elements of the samevalue as X) in A, then X is called one of the most frequent numbers of A. Now asequence of numbers A of length L is given, and it is assumed that there is anumber X which has more than L / 2 instances in A. Apparently X is the only onemost frequent number of A. Could you find out X with a very limited memory?

Input

Input containsmultiple test cases. Each test case there is one line, which starts with anumber L (1 <= L <= 250000), followed by L numbers (-2^31 ~ 2^31-1).Adjacent numbers is separated by a blank space.

Output

There is one line foreach test case, which is the only one most frequent number X.

Sample Input

5 2 1 2 3 2

8 3 3 4 4 4 4 3 4

扫描二维码关注公众号，回复： 812756 查看本文章

Sample Output

2

4

0x01 题意

求序列众数。

0x02 分析

第一次是在啊哈算法最后一章学到的，书中给出复杂度逐次降低的算法。

大致意思是，因为最多的数超过一半，例如第一个sample中，2众数，可以将每个2都分配给不同的其它元素，最后2还是有剩余，就像是同归于尽一样，最后剩下的就是其中最多的元素了。利用这个我们可以定义一个当前数和当前数的个数下一个比，相同则加一，不同则减一知道个数为0时，当前数变成下一个，知道输入结束，存活下来的就是众数。

0x03 Code

一遍AC：

//ZOJ-2132
//2017-06-18
#include<bits/stdc++.h>
int n, num;
int main() {
	while(scanf("%d%d", &n, &num)!=EOF) {
		int cnt = 1;
		for (int i = 2; i <= n; i++) {
			int tmp;
			scanf("%d", &tmp);
			if (cnt == 0) {
				cnt = 1;
				num = tmp;
				continue;
			}
			if (tmp == num) cnt++;
			else cnt--;
		}
		printf("%d\n", num);
	}
	return 0;
}