ZOJ 2132 The Most Frequent Number 贪心

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.

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Output

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

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Sample Input

5 2 1 2 3 28 3 3 4 4 4 4 3 4

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Sample Output

24

很经典的贪心题，最多的数比剩余的数多，所以直接贪心

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int n;int main(){    while(~scanf("%d",&n))    {        int t,a;        int sum=0;        for(int i=0;i<n;i++)        {            scanf("%d",&a);            if(i==0)            {                sum++;                t=a;            }            else            {                if(a==t)    sum++;                else                {                    sum--;                }                if(sum<0)                {                    t=a;                    sum=1;                }            }        }        printf("%d\n",t);    }}