ZOJ 2132 The Most Frequent Number (贪心)

题意：给定一个序列，里面有一个数字出现了超过 n / 2，问你是哪个数字，但是内存只有 1 M。

析：首先不能开数组，其实也是可以的了，后台数据没有那么大，每次申请内存就可以过了。正解应该是贪心，模拟一个栈，因为答案肯定出现次数比其他所有数字的出现次数还多，所以每次和栈顶的元素比较，如果相同，就放到栈里，否则删除栈顶元素，这样最后剩下的肯定是出现次数最多的元素，我们可以用两个变量来模拟这个栈。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 1e5 + 5;
const int maxm = 700 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int main(){
  while(scanf("%d", &n) == 1){
    int cnt = 0, t = 0;
    for(int i = 0; i < n; ++i){
      scanf("%d", &m);
      if(cnt == 0){
        t = m;  ++cnt;
      }
      else{
        if(m == t)  ++cnt;
        else  --cnt;
        if(cnt < 0)  cnt = 1, t = m;
      }
    }
    printf("%d\n", t);
  }
  return 0;
}