HUST 1343 Reverse Number(哈理工 亞洲區選拔賽前練習賽)

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G - Reverse NumberTime Limit:1000MS    Memory Limit:131072KB    64bit IO Format:%lld & %lluSubmitStatusPracticeHUST 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 13 2 15 25 1 4 3 2

Sample Output

Case #1: 2Case #2: 5 先用樹狀數組求出逆序數。因為每一次交換可以增加或著減少一對逆序數，假設有m對逆序數，我們交換n對，那麼這n對我們讓他每次都減少一對逆序數，交換n次後 還有m - n對逆序。注意題目中k的取值，如果求出的逆序數大於k，那麼可以直接得出結果 res - k ，如果小於k，此時就要注意數字串中是否有重複的，如果沒有那麼當交換res - k次後此時逆序數恰好為0， 剩餘交換次數為k - res，如果k - res為偶數，那麼我們可以重複交換同一對此時逆序數還為0，如果為奇數，此時只能結果為1。  如果字串中有重複的那麼可以交換那兩個重複的數，此時無論是奇數還是偶數，結果並不影響最小逆序對總數。
/*=============================================================================##      Author: liangshu - cbam ##      QQ : 756029571 ##      School : 哈爾濱理工大學 ##      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 100010struct 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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HUST 1343 Reverse Number(哈理工 亞洲區選拔賽前練習賽)