uva 11991 Easy Problem from Rujia Liu? Data Structure

// uva 11991 Easy Problem from Rujia Liu?// 给一个包含n个数的数组，需要回答若干个询问，每次询问两个整数// k和v，输出从左到右第k个v的下标//// 本题因为n比较大，所以直接开二维数组是不现实的// 如果直接用vector的话，也是会有大量的浪费// 所以可以先离散化一下，找到一共有多少个不同的数// 再用vector动态数组就可以搞定了//// 看着书上的map写的挺精彩的，我就习用了下来//// 非常精彩，继续练吧。。。#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cfloat>#include <climits>#include <cmath>#include <complex>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <vector>#define ceil(a,b) (((a)+(b)-1)/(b))#define endl '\n'#define gcd __gcd#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))#define popCount __builtin_popcountlltypedef long long ll;using namespace std;const int MOD = 1000000007;const long double PI = acos(-1.L);template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }template<class T> inline T lowBit(const T& x) { return x&-x; }template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }map<int,vector<int> > mp;int n,m;void init(){mp.clear();int x;for (int i=0;i<n;i++){scanf("%d",&x);if (!mp.count(x))mp[x] = vector<int> ();mp[x].push_back(i+1);}for (int i=0;i<m;i++){int v,k;scanf("%d%d",&k,&v);if (!mp.count(v) || mp[v].size()<k){puts("0");}elseprintf("%d\n",mp[v][k-1]);}}int main() {//freopen("G:\\Code\\1.txt","r",stdin);while(scanf("%d%d",&n,&m)!=EOF){init();//solve();}return 0;}