hdu 5199 Gunner（hash）

Gunner

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1727    Accepted Submission(s): 756

Problem Description

Long long ago, there is a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i−th bird stands on the top of the i−th tree. The trees stand in straight line from left to the right. Every tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the bird which stands in the tree with height H will falls.
Jack will shot many times, he wants to know how many birds fall during each shot.

a bullet can hit many birds, as long as they stand on the top of the tree with height of H.

 

Input

There are multiple test cases (about 5), every case gives n,m in the first line, n indicates there are n trees and n birds, m means Jack will shot m times.

In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.

In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.

Please process to the end of file.

[Technical Specification]

1≤n,m≤1000000(106)

1≤h[i],q[i]≤1000000000(109)

All inputs are integers.

 

Output

For each q[i], output an integer in a single line indicates the number of birds Jack shot down.

 

Sample Input

4 31 2 3 41 1 4

 

Sample Output

101
Hint
Huge input, fast IO is recommended.

题意：有n课树，每棵树上有一只鸟站在树的顶端，给出每棵树的高度

猎人依次从m个高度射出一颗子弹，问每次能射死几只鸟

思路：用hash存下树的高度，然后每次取找有多少个一样的即可

用map更方便

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <map>using namespace std;#define N 1000005map<int,int>ma;int main(){    int n,m,t;    while(~scanf("%d %d",&n,&m))    {        ma.clear();        for(int i=1;i<=n;i++)            {                scanf("%d",&t);                ma[t]++;            }        for(int i=1;i<=m;i++)        {            scanf("%d",&t);            printf("%d\n",ma[t]);            ma[t]=0;        }    }    return 0;}