Though Rujia Liu usually sets hard problems for contests (for example, regional contests like Xi‘an 2006, Beijing 2007 and Wuhan 2009, or UVa OJ contests like Rujia Liu‘s Presents 1 and 2), he occasionally sets easy problem (for example, ‘the Coco-Cola Store‘ in UVa OJ), to encourage more people to solve his problems :D
Given an array, your task is to find the k-th occurrence (from left to right) of an integer v. To make the problem more difficult (and interesting!), you‘ll have to answer m such queries.
There are several test cases. The first line of each test case contains two integers n, m(1<=n,m<=100,000), the number of elements in the array, and the number of queries. The next line contains n positive integers not larger than 1,000,000. Each of the following m lines contains two integer k and v (1<=k<=n, 1<=v<=1,000,000). The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.
For each query, print the 1-based location of the occurrence. If there is no such element, output 0 instead.
8 4 1 3 2 2 4 3 2 1 1 3 2 4 3 2 4 2
2 0 7 0
Note: Please make sure to test your program with the gift I/O files before submitting!
哈希的思想,本来想用结构体,但数组太大开不下,所以用map,一维用来存储对应的v,二维存储对应的第k个v的下标
#include<iostream> #include<cstring> #include<cstdio> #include<string> #include<algorithm> #include<vector> #include<map> using namespace std; map<int,vector<int> >mp; map<int,vector<int> >::iterator it; int a; int n,m; int k,v; int main() { freopen("in.txt","r",stdin); while(~scanf("%d%d",&n,&m)){ mp.clear(); for(int i=1;i<=n;i++){ scanf("%d",&a); /*if(mp.count(a)==0){ mp[a]=vector<int> {}; }*/ mp[a].push_back(i); } for(int i=0;i<m;i++){ scanf("%d%d",&k,&v); if(mp.count(v)==0||mp[v].size()<k)puts("0"); else { printf("%d\n",mp[v][k-1]); } } } }
uva Easy Problem from Rujia Liu?
原文地址:http://blog.csdn.net/u013497977/article/details/44001799