uva 11991 Easy Problem from Rujia Liu?

2019-11-14 11:25:21

字體：大中小

來源：轉載

供稿：網友

原題： 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. Input 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). Output For each query, print the 1-based location of the occurrence. If there is no such element, output ‘0’instead. Sample Input 8 4 1 3 2 2 4 3 2 1 1 3 2 4 3 2 4 2 Sample Output 2 0 7 0

中文： 給你n個數，然后給你m個查詢，每個查詢包含兩個數k和v問你第k個v的小標是多少？

#include <bits/stdc++.h>using namespace std;unordered_map<int,vector<int>> miv;int n,m;int main(){ ios::sync_with_stdio(false); while(cin>>n>>m) { miv.clear(); for(int i=1;i<=n;i++) { int res; cin>>res; miv[res].push_back(i); } for(int i=1;i<=m;i++) { int k,v; cin>>k>>v; if(miv.find(v)==miv.end()||miv[v].size()<k) cout<<0<<endl; else { cout<<miv[v][k-1]<<endl; } } } return 0;}

思路： 這題的名起的，不秒殺都不好意思~~ 思路見下面的圖這里寫圖片描述長方格里面的數是輸入的所有數，每個方格里面的數下面的鏈是按順序存儲的下標值。