暑假集训第三周周三赛 STL E - Gunner 射鸟

E - Gunner

Time Limit:4000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5199

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] 

$1 \leq n,m \leq 1000000(10^{6})$ 

$1 \leq h[i],q[i] \leq 1000000000(10^{9})$ 

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 

分析：

本题用到一个find（）函数，它也是遍历，但它只能遍历找到一个数据，而迭代器是找到所有数据

比如 3 3 find只能输出一个，而迭代器可以输出两个

1234567891011121314151617181920212223242526272829

#include<stdio.h>#include<map>using namespace std;int main(){    int m,n,a,b,i;    while(scanf("%d %d",&m,&n)!=EOF)    {        map< int,int > p;        for(i=0;i<m;i++)        {            scanf("%d",&a);            p[a]++;        }        for(i=0;i<n;i++)        {            scanf("%d",&b);            if(p.find(b)!=p.end())            {                printf("%d\n",p[b]);                p.erase(b);            }            else                printf("0\n");        }    }    return 0;}