zoukankan      html  css  js  c++  java
  • Codeforces Round #261 (Div. 2) D

    D. Pashmak and Parmida's problem
    time limit per test
    3 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partner to be clever too (although he's not)! Parmida has prepared the following test problem for Pashmak.

    There is a sequence a that consists of n integers a1, a2, ..., an. Let's denote f(l, r, x) the number of indices k such that: l ≤ k ≤ r andak = x. His task is to calculate the number of pairs of indicies i, j (1 ≤ i < j ≤ n) such that f(1, i, ai) > f(j, n, aj).

    Help Pashmak with the test.

    Input

    The first line of the input contains an integer n (1 ≤ n ≤ 106). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

    Output

    Print a single integer — the answer to the problem.

    sl :很傻的数据结构题,直接求下逆序数就好了。

    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <map>
    #include <algorithm>
    using namespace std;
    const int MAX= 1e6+10;
    int a[MAX],C[MAX];
    map<int,int> hash1,hash2,hash3,cnt1,cnt2;
    int lowbit(int x) {
        return x&-x;
    }
    void add(int x,int val) {
        for(int i=x;i<MAX;i+=lowbit(i)) {
            C[i]+=val;
        }
    }
    int sum(int x) {
        int res=0;
        for(int i=x;i>0;i-=lowbit(i)) res+=C[i];
        return res;
    }
    int main() {
        int n;
        while(scanf("%d",&n)==1) {

            memset(C,0,sizeof(C));
            for(int i=1;i<=n;i++) {
                scanf("%d",&a[i]);
                cnt1[a[i]]++;
                hash1[i]=cnt1[a[i]];
            }
            long long ans=0;
            for(int i=n;i>=1;i--) {

                cnt2[a[i]]++;
                hash2[i]=cnt2[a[i]];
                add(hash2[i],1);
                if(i-1>=1)
                ans+=sum(hash1[i-1]-1);
            }
            printf("%I64d ",ans);
        }
        return 0;

    } 

  • 相关阅读:
    Leetcode Valid Sudoku
    Leetcode Surrounded Regions
    LeetCode Sqrt
    LeetCode POW
    LeetCode Next Permutation
    ACK-Ackermann, 阿克曼函数
    再不懂时序就 OUT 啦!,DBengine 排名第一时序数据库,阿里云数据库 InfluxDB 正式商业化!
    阿里云提供全托管 ZooKeeper
    性能压测中的SLA,你知道吗?
    第一个入驻阿里云自营心选商城,如今它已经是营收过亿的SaaS独角兽
  • 原文地址:https://www.cnblogs.com/acvc/p/3916180.html
Copyright © 2011-2022 走看看