11991

?Easy Problem from Rujia Liu?

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.
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

/* Author:2486 Memory: 0 KB Time: 123 MS Language: C++ 4.8.2 Result: Accepted */ //通過<sstream>中的stringstream進行字符串處理 #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; }

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