二分查找 BestCoder Round #36 ($) Gunner

 

题目传送门

 1 /*  2  题意：问值为x的个数有几个，第二次查询就是0  3  lower/upper_bound ()函数的使用，map也可过，hash方法不会  4 */  5 #include <cstdio>  6 #include <cmath>  7 #include <cstring>  8 #include <algorithm>  9 #include <iostream> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <queue> 14 using namespace std; 15 16 const int MAXN = 1e6 + 10; 17 const int INF = 0x3f3f3f3f; 18 int a[MAXN], b[MAXN]; 19 int used[MAXN]; 20 int n, m; 21 22 inline int read(void) 23 { 24 int x = 0, f = 1; char ch = getchar (); 25 while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar ();} 26 while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar ();} 27 return f * x; 28 } 29 30 int main(void) //HDOJ 5199 Gunner 31 { 32 //freopen ("B.in", "r", stdin); 33 34 while (scanf ("%d%d", &n, &m) == 2) 35  { 36 memset (used, 0, sizeof (used)); 37 for (int i=1; i<=n; ++i) a[i] = read (); 38 for (int i=1; i<=m; ++i) b[i] = read (); 39 40 sort (a+1, a+1+n); 41 for (int i=1; i<=m; ++i) 42  { 43 int x = upper_bound (a+1, a+1+n, b[i]) - a; 44 int y = lower_bound (a+1, a+1+n, b[i]) - a; 45 if (used[x]) 46  { 47 puts ("0"); continue; 48  } 49 used[x] = 1; 50 printf ("%d\n", x - y); 51  } 52  } 53 54 return 0; 55 }

 1 #include <cstdio>  2 #include <iostream>  3 #include <cstring>  4 #include <map>  5 using namespace std;  6  7 const int MAXN = 1e6 + 10;  8 const int INF = 0x3f3f3f3f;  9 10 inline int read(void) 11 { 12 int x = 0, f = 1; char ch = getchar (); 13 while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar ();} 14 while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar ();} 15 return f * x; 16 } 17 int main(void) 18 { 19 //freopen ("B.in", "r", stdin); 20 21 int n, m; 22 while (scanf ("%d%d", &n, &m) == 2) 23  { 24 map<int, int> ma; int x; 25 while (n--) 26  { 27 x = read (); ma[x]++; 28  } 29 map<int, int>::iterator it; 30 while (m--) 31  { 32 x = read (); 33 it = ma.find (x); 34 if (it == ma.end ()) 35 puts ("0"); 36 else 37  { 38 printf ("%d\n", it->second); 39 it->second = 0; 40  } 41  } 42  } 43 44 return 0; 45 }

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