hdu 5199 Gunner【水题】【STL应用】【fast IO】

Gunner

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

Total Submission(s): 1637 Accepted Submission(s): 705



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

HintHuge input, fast IO is recommended.

这里数据量比较大，建议用fast IO读入，但是好像用scanf就行，这里既然用了，就贴出来。

卡内存的一个题、10^9，一维数组不能开那么大，用map直接搞定就行了:

#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
int read()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        map<int,int >a;
        for(int i=0;i<n;i++)
        {
            int k;
            k=read();
            a[k]++;
        }
        while(m--)
        {
            int k;
            k=read();
            printf("%d\n",a[k]);
            a[k]=0;
        }
    }
}