【題解】 HDU 5199 Gunner

目錄

 

題目描述

題意分析

AC代碼

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題目描述

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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 3

1 2 3 4

1 1 4

Sample Output

1

0

1

 

題意分析

題意：N棵 高H 的樹頂上有一隻鳥 ，這個獵人接下來會發射 M次 高爲H的子彈。掃到了，鳥兒就die了

           問你每次打落幾隻鳥。數據十分大，用CIN會直接TE掉。MAP內自建一顆紅黑樹，速度還是很快的

 

AC代碼

#include <iostream>
#include <map>
#define MAXN 1000005
using namespace std;

map <long long,long long> s;

int main()
{
    long long n,m,i,p;
    while(~scanf("%lld%lld",&n,&m))
    {
        s.clear();
        for(i=0;i<n;i++)
        {
            scanf("%lld",&p);
            s[p]++;
        }
        for(i=0;i<m;i++)
        {
            scanf("%lld",&p);
            printf("%lld",s[p]);
            if(s[p]>0)
            {
                s[p]=0;
            }
            printf("\n");
        }
    }
    return 0;
}