bestcoder#36 1002 哈希

bestcoder#36 1002 哈希

Gunner

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 726    Accepted Submission(s): 326

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

Hint

Huge input, fast IO is recommended.

 

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<ctype.h>

using namespace std;

const int maxn=1000100;
const int HM=1000100;

int n,m;
struct Hash
{
    int date;
    int cnt;
    Hash *next;
};
Hash H[maxn];

inline int read()
{
    char c=getchar();
    while(!isdigit(c)) c=getchar();
    int f=c-'0';
    while(isdigit(c=getchar())) f=f*10+c-'0';
    return f;
}

int h(int x)
{
    return x%HM;
}

void insert(Hash*H,int date)
{
    int key=h(date);
    Hash *pre=&H[key],*p=pre->next;
    while(p!=NULL){
        if(p->date==date){
            p->cnt++;
            return;
        }
        pre=p;
        p=p->next;
    }
    p=(Hash*)malloc(sizeof(Hash));
    pre->next=p;
    p->next=NULL;
    p->date=date;
    p->cnt=1;
}

void find_del_output(Hash*H,int date)
{
    int key=h(date);
    Hash *pre=&H[key],*p=pre->next;
    while(p!=NULL){
        if(p->date==date){
            printf("%d
",p->cnt);
            pre->next=p->next;///删掉结点
            free(p);
            return;
        }
        pre=p;
        p=p->next;
    }
    puts("0");
}

int main()
{
    while(cin>>n>>m){
        memset(H,0,sizeof(H));
        while(n--){
            int h=read();
            insert(H,h);
        }
        while(m--){
            int q=read();
            find_del_output(H,q);
        }
    }
    return 0;
}