HUST 1343 Reverse Number(哈理工 亚洲区选拔赛前练习赛)

G - Reverse Number

Time Limit:1000MS    Memory Limit:131072KB    64bit IO Format:%lld & %llu

SubmitStatusPracticeHUST
1347

Description

Given a non-negative integer sequence A with length N, you can exchange two adjacent numbers each time. After K exchanging operations, what’s the minimum reverse number the sequence can achieve? The reverse number of a sequence is the number
of pairs (i, j) such that i < j and Ai > Aj

Input

There are multiple cases. For each case, first line contains two numbers: N, K 2<=N<=100000, 0 <= K < 2^60 Second line contains N non-negative numbers, each of which not greater than 2^30

Output

Minimum reverse number you can get after K exchanging operations.

Sample Input

3 1
3 2 1
5 2
5 1 4 3 2

Sample Output

Case #1: 2
Case #2: 5
 先用树状数组求出逆序数。因为每一次交换可以增加或着减少一对逆序数，假设有m对逆序数，我们交换n对，那么这n对我们让他每次都减少一对逆序数，交换n次后 还有m - n对
逆序。注意题目中k的取值，如果求出的逆序数大于k，那么可以直接得出结果 res - k ，如果小于k，此时就要注意数字串中是否有重复的，如果没有那么当交换res - k次后
此时逆序数恰好为0， 剩余交换次数为k - res，如果k - res为偶数，那么我们可以重复交换同一对此时逆序数还为0，如果为奇数，此时只能结果为1。  如果字串中有重复的
那么可以交换那两个重复的数，此时无论是奇数还是偶数，结果并不影响最小逆序对总数。

/*=============================================================================
#
#      Author: liangshu - cbam
#
#      QQ : 756029571
#
#      School : 哈尔滨理工大学
#
#      Last modified: 2015-08-30 22:32
#
#     Filename: A.cpp
#
#     Description:
#        The people who are crazy enough to think they can change the world, are the ones who do !
=============================================================================*/
#
#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define maxn 100010
struct node
{
    int v,id;
} s[maxn];
int c[maxn],n;
typedef long long ll;
ll res;

bool cmp(node x,node y)
{
    return ((x.v>y.v) || ((x.v==y.v)&&(x.id>y.id)));
}

int Lowbit(int x)
{
    return x&(x^(x-1));
}
ll Getsum(int pos)
{
    ll ret = 0LL;
    while(pos>0)
    {
        ret+=c[pos];
        pos -= Lowbit(pos);
    }
    return ret;
}
ll update(int pos)
{
    while(pos<=n)
    {
        c[pos]++;
        pos+=Lowbit(pos);
    }
}

int main()
{
    int k,x;
    int cs = 1;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        set<int>cnt;
        memset(c,0,sizeof(c));
        res = 0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&s[i].v);
            cnt.insert(s[i].v);
            s[i].id = i;
        }
        sort(s+1,s+n+1,cmp);
        for(int i=1; i<=n; i++)
        {
            res += Getsum(s[i].id);
            update(s[i].id);
        }
        if((res-k)>=0)
            printf("Case #%d: %lld\n",cs ++,res-k);
        else
        {
            if(cnt.size() < n)
            {
                printf("Case #%d: 0\n",cs ++);
            }
            else
            {
                if(abs(res-k)%2) printf("Case #%d: 1\n",cs ++);
                else
                    printf("Case #%d: 0\n",cs ++);
            }
        }
    }
    return 0;
}

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