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uva——11991——Problem E

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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). The size of input file does not exceed 5MB. 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 Output for the Sample Input 2 0 7 0 题目:给一串数,让你输出第k次出现的数字v的下标(下标从1开始) 刚开始想简单了,用数组存会超时 下面使用容器vector方法: #include <iostream> #include <cstdio> #include <vector> using namespace std; const int MAXN=1000000+10; vector<int>data[MAXN]; int main() { int n,m; int temp; int k,v; while(scanf("%d%d",&n,&m)!=EOF) { for(int i=0;i<MAXN;i++) data[i].clear(); for(int i=1;i<=n;i++) { scanf("%d",&temp); data[temp].push_back(i); } for(int i=0;i<m;i++) { scanf("%d%d",&k,&v); if(data[v].empty()||k>data[v].size()) { printf("0\n"); } else { printf("%d\n",data[v][k-1]); } } } return 0; }

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