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。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <cassert> 24 #include <cstdio> 25 #include <bitset> 26 #include <vector> 27 #include <deque> 28 #include <queue> 29 #include <stack> 30 #include <ctime> 31 #include <set> 32 #include <map> 33 #include <cmath> 34 using namespace std; 35 #define fr first 36 #define sc second 37 #define cl clear 38 #define BUG puts("here!!!") 39 #define W(a) while(a--) 40 #define pb(a) push_back(a) 41 #define Rint(a) scanf("%d", &a) 42 #define Rll(a) scanf("%I64d", &a) 43 #define Rs(a) scanf("%s", a) 44 #define Cin(a) cin >> a 45 #define FRead() freopen("in", "r", stdin) 46 #define FWrite() freopen("out", "w", stdout) 47 #define Rep(i, len) for(int i = 0; i < (len); i++) 48 #define For(i, a, len) for(int i = (a); i < (len); i++) 49 #define Cls(a) memset((a), 0, sizeof(a)) 50 #define Clr(a, x) memset((a), (x), sizeof(a)) 51 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 52 #define lrt rt << 1 53 #define rrt rt << 1 | 1 54 #define pi 3.14159265359 55 #define RT return 56 #define lowbit(x) x & (-x) 57 #define onenum(x) __builtin_popcount(x) 58 typedef long long LL; 59 typedef long double LD; 60 typedef unsigned long long ULL; 61 typedef pair<int, int> pii; 62 typedef pair<string, int> psi; 63 typedef pair<LL, LL> pll; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<LL> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 typedef struct Node { 71 char p[5]; 72 int c; 73 }Node; 74 75 const int maxn = 1100; 76 int n; 77 char qx[5]; 78 Node k[maxn]; 79 80 int main() { 81 // FRead(); 82 while(~Rint(n)) { 83 Cls(k); 84 For(i, 1, n+1) { 85 Rs(qx); Rs(k[i].p); Rint(k[i].c); 86 k[i].c <<= 2; 87 } 88 int ret = 0; 89 For(x, -1005, 4005) { 90 int cur = 0; 91 For(i, 1, n+1) { 92 int len = strlen(k[i].p); 93 if(k[i].p[0] == '<' && len == 1) 94 if(x < k[i].c) cur++; 95 if(k[i].p[0] == '<' && k[i].p[1] == '=') 96 if(x <= k[i].c) cur++; 97 if(k[i].p[0] == '=' && len == 1) 98 if(x == k[i].c) cur++; 99 if(k[i].p[0] == '>' && len == 1) 100 if(x > k[i].c) cur++; 101 if(k[i].p[0] == '>' && k[i].p[1] == '=') 102 if(x >= k[i].c) cur++; 103 } 104 ret = max(ret, cur); 105 } 106 printf("%d ", ret); 107 } 108 RT 0; 109 }