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HDU 5199

时间:2015-05-28 10:54:03      阅读:153      评论:0      收藏:0      [点我收藏+]

标签:acm   hdu   

Gunner

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1323    Accepted Submission(s): 586


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]

1n,m1000000(106)

1h[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
Hint
Huge input, fast IO is recommended.
 

Source
#include <cstdio>
#include <iostream>
#include <map>

using namespace std;
#define maxn 1000000 + 10

int n, m;
map<int, int> M;

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        int t;
        M.clear();
        for(int i=0; i<n; i++)
        {
            scanf("%d", &t);
            M[t]++;
        }
        for(int i=0; i<m; i++)
        {
            scanf("%d", &t);
            if(M[t])
            {
                printf("%d\n", M[t]);
                M.erase(t);
            }
            else
                printf("0\n");
        }
    }
    return 0;
}


HDU 5199

标签:acm   hdu   

原文地址:http://blog.csdn.net/dojintian/article/details/46119205

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