[HIHO1223]不等式（离散化，枚举）

题目链接：http://hihocoder.com/problemset/problem/1223

这题不难，难点在于小数的处理。可以0.5为步长枚举，也可以扩大偶数倍枚举。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 typedef struct Node {
     char p[];
     int c;
 }Node;

 const int maxn = ;
 int n;
 char qx[];
 Node k[maxn];

 int main() {
     // FRead();
     while(~Rint(n)) {
         Cls(k);
         For(i, , n+) {
             Rs(qx); Rs(k[i].p); Rint(k[i].c);
             k[i].c <<= ;
         }
         int ret = ;
         For(x, -, ) {
             int cur = ;
             For(i, , n+) {
                 int len = strlen(k[i].p);
                 if(k[i].p[] == '<' && len == )
                     if(x < k[i].c) cur++;
                 if(k[i].p[] == '<' && k[i].p[] == '=')
                     if(x <= k[i].c) cur++;
                 if(k[i].p[] == '=' && len == )
                     if(x == k[i].c) cur++;
                 if(k[i].p[] == '>' && len == )
                     if(x > k[i].c) cur++;
                 if(k[i].p[] == '>' && k[i].p[] == '=')
                     if(x >= k[i].c) cur++;
             }
             ret = max(ret, cur);
         }
         printf("%d\n", ret);
     }
     RT ;
 }