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Easy Problem from Rujia Liu?
Though Rujia Liu usually sets hard problems for contests (for example, regionalcontests like Xi’an 2006, Beijing 2007 and Wuhan 2009, or UVa OJ contests likeRujia Liu’s Presents 1 and 2), he occasionally sets easy problem (for example, ‘theCoco-Cola Store’ in UVa OJ), to encourage more people to solve his problems :DGiven an array, your task is to find the k-th occurrence (from left to right) of an integer v. To makethe problem more difficult (and interesting!), you’ll have to answer m such queries.InputThere 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 npositive 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).OutputFor each query, print the 1-based location of the occurrence. If there is no such element, output ‘0’instead.Sample Input8 41 3 2 2 4 3 2 11 32 43 24 2Sample Output207
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/*Author:2486Memory: 0 KBTime: 123 MSLanguage: C++ 4.8.2Result: Accepted*///通过<sstream>中的stringstream进行字符串处理#include <cstdio>#include <cstring>#include <algorithm>#include <iterator>#include <vector>using namespace std;const int maxn=100000+5;struct ins {int id,val;ins(int val,int id):id(id),val(val) {}bool operator < (const ins & a)const {if(val!=a.val)return val<a.val;return id<a.id;}};vector<ins>buf;int k,v,n,m;int main() {#ifndef ONLINE_JUDGEfreopen("D:imput.txt","r",stdin);#endif // ONLINE_JUDGEwhile(~scanf("%d%d",&n,&m)) {int val;buf.clear();for(int i=0; i<n; i++) {scanf("%d",&val);buf.push_back(ins(val,i+1));}sort(buf.begin(),buf.end());for(int i=0; i<m; i++) {scanf("%d%d",&k,&v);ins s(v,0);//printf("[][][][][][]\n");vector<ins>::iterator buf1=lower_bound(buf.begin(),buf.end(),s);if(buf1!=buf.end()&&buf1->val==v&&k!=0) {buf1+=(k-1);if(buf1->val==v)printf("%d\n",buf1->id);else printf("0\n");} else printf("0\n");}}return 0;}
，自然而然不想去因为别人的努力而努力，
