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The kth number
Time Limit: 12000/6000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)SubmitStatisticNext ProblemProblem Description
Do you still remember the Daming Lake‘s k‘th number? Let me take you back and recall that wonderful memory.
Given a sequence A with length of n,and m querys.Every query is defined by three integer(l,r,k).For each query,please find the kth biggest frequency in interval [l,r].
Frequency of a number x in [l,r] can be defined by this code:
123456int
FrequencyOfX = 0;
for
(
int
i = l; i <= r; i ++) {
if
(a[i]==X) {
FrequencyOfX ++;
}
}
Input
First line is a integer T,the test cases.
For each case:
First line contains two integers n and m.
Second line contains n integers a1,a2,a3....an.
Then next m lines,each line contain three integers l,r,k.T<=12
1<=n,m,ai<=100000
1<=l<=r<=n
1<=k
data promise that for each query(l,r,k),the kind of number in interval [l,r] is at least k.Output
for every query,output a integer in a line.Sample Input
1 6 3 13 14 15 13 14 13 1 6 3 1 6 1 3 5 2Sample Output
1 3 1Source
zhangmingmingManager
1 /* 2 * this code is made by Stomach_ache 3 * Problem: 1108 4 * Verdict: Accepted 5 * Submission Date: 2014-09-04 21:32:52 6 * Time: 1320MS 7 * Memory: 4248KB 8 */ 9 #include <stdio.h> 10 #include <string.h> 11 #include <algorithm> 12 using namespace std; 13 /*Let‘s fight!!!*/ 14 15 const int Sqrt = 333; 16 const int MAX_N = 101000; 17 int a[MAX_N], ans[MAX_N], freq[MAX_N], cnt[MAX_N]; 18 int ll[MAX_N], rr[MAX_N], kk[MAX_N], idx[MAX_N], n, m; 19 20 bool cmp (int a, int b) { 21 if (ll[a]/Sqrt == ll[b]/Sqrt) return rr[a] < rr[b]; 22 return ll[a] < ll[b]; 23 } 24 25 int query(int k) { 26 int lb = 1, ub = 100001; 27 while (ub - lb > 1) { 28 int mid = (lb + ub) / 2; 29 if (freq[mid] >= k) lb = mid; 30 else ub = mid; 31 } 32 return lb; 33 } 34 35 int main() { 36 int T; 37 scanf("%d", &T); 38 while (T--) { 39 scanf("%d%d", &n, &m); 40 for (int i = 0; i < n; i++) scanf("%d", a+i); 41 for (int i = 0; i < m; i++) { 42 scanf("%d%d%d", ll+i, rr+i, kk+i); 43 idx[i] = i; ll[i]--; rr[i]--; 44 } 45 46 sort(idx, idx + m, cmp); 47 memset(freq, 0, sizeof(freq)); 48 memset(cnt, 0, sizeof(cnt)); 49 50 int cl = 0, cr = -1; 51 for (int i = 0; i < m; i++) { 52 int l = ll[idx[i]], r = rr[idx[i]], k = kk[idx[i]]; 53 while (cr < r) { freq[++cnt[a[++cr]]] ++; } 54 while (l < cl) { freq[++cnt[a[--cl]]] ++; } 55 while (r < cr) { freq[cnt[a[cr--]]--] --; } 56 while (cl < l) { freq[cnt[a[cl++]]--] --; } 57 ans[idx[i]] = query(k); 58 } 59 60 for (int i = 0; i < m; i++) printf("%d\n", ans[i]); 61 } 62 63 return 0; 64 }
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原文地址:http://www.cnblogs.com/Stomach-ache/p/3956935.html