zoj2132-The Most Frequent Number
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2132
Seven (actually six) problems may be somewhat few for a contest. But I am really unable to devise another problem related to Fantasy Game Series. So I make up an very easy problem as the closing problem for this contest.
Given a sequence of numbers A, for a number X if it has the most instances (elements of the same value as X) in A, then X is called one of the most frequent numbers of A. Now a sequence of numbers A of length L is given, and it is assumed that there is a number X which has more than L / 2 instances in A. Apparently X is the only one most frequent number of A. Could you find out X with a very limited memory?
Input
Input contains multiple test cases. Each test case there is one line, which starts with a number 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 for each 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
Sample Output
2 4
思路:题意很简单,求一个数列里出现次数最多的那个数字,这个数字的个数是大于数列长度的1/2的。这里采用O(1)算法,利用大于1/2这个条件。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 3 int main() 4 { 5 int n,ans; 6 while(~scanf("%d",&n)) 7 { 8 int m,cnt=0; 9 for(int i=0;i<n;i++) 10 { 11 scanf("%d",&m); 12 if(cnt==0) ans=m,cnt++; 13 else if(ans==m) cnt++; 14 else cnt--; 15 } 16 printf("%d\n",ans); 17 } 18 return 0; 19 }