HDU 5233 杂题

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Gunner II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1499 Accepted Submission(s): 569

Problem Description

Long long ago, there was 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 nearest bird which stands in the tree with height H will falls.

Jack will shot many times, he wants to know which bird will fall during each shot.

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]

All input items are integers.

1<=n,m<=100000(10^5)

1<=h[i],q[i]<=1000000000(10^9)

Output

For each q[i], output an integer in a single line indicates the id of bird Jack shots down. If Jack can’t shot any bird, just output -1.

The id starts from 1.

Sample Input

5 51 2 3 4 11 3 1 4 2

Sample Output

#include <cstdio>#include <string.h>#include<cstdio>#include <map>using namespace std;int kfc[100005];// kfc[x]表示下一个高度和序号x的树一样高的序号 int last[100005];//last[a[x]]最后一个序号  map<int,int> a;//a[x]表示高度为x的第一个序号 int main(){int m,n,q,k,gao,t;while(scanf("%d%d",&n,&m)==2){a.clear();memset(kfc,0,sizeof kfc);for(int i=1;i<=n;i++){scanf("%d",&gao);if(a[gao]==0){a[gao]=i;}else{kfc[last[a[gao]]]=i;}last[a[gao]]=i;}for(int i=0;i<m;i++){scanf("%d",&q);if(a[q]==0)puts("-1");else{printf("%d\n",a[q]);a[q]=kfc[a[q]];}}}return 0;}13542

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