首页 > 代码库 > Codeforces Round #261 (Div. 2)459D. Pashmak and Parmida's problem（求逆序数对）

Codeforces Round #261 (Div. 2)459D. Pashmak and Parmida's problem（求逆序数对）

2024-07-18 00:11:53   235人阅读

题目链接：http://codeforces.com/contest/459/problem/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.

Sample test(s)
input
7
1 2 1 1 2 2 1
output
8
input
3
1 1 1
output
1
input
5
1 2 3 4 5
output
0


思路：用map预处理出 f(1, i, a[i]) 和 f(j, n, a[j]) ，再求逆序数对！


代码如下：

#include <cstdio>
#include <map>
using namespace std;
typedef long long LL;
const int Maxn = 1000017;
int n;
int a[Maxn],c[Maxn];
int f[Maxn];
LL res;
map <int, int> freq;
int Lowbit(int x) //2^k
{
    return x&(-x);
}

void update(int i, int x)//i点增量为x
{
    while(i <= n)
    {
        c[i] += x;
        i += Lowbit(i);
    }
}
int sum(int x)//区间求和 [1,x]
{
    int sum=0;
    while(x>0)
    {
        sum+=c[x];
        x-=Lowbit(x);
    }
    return sum;
}
int main()
{
    while(~scanf("%d", &n))
    {
        for (int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        for (int i = n - 1; i >= 0; i--)
            f[i] = ++freq[a[i]];
        freq.clear();
        //树状数组求逆序
        for (int i = 0; i < n; i++)
        {
            //res += sum(f[i] + 1);
            res += i - sum(f[i]);//逆序数个数
            update(++freq[a[i]],1);
        }
        printf("%I64d\n", res);
    }
    return 0;
}


贴一个讲解比较详细的链接：http://www.cnblogs.com/bfshm/p/3916799.html


联系
我们