hdu4911：Inversion

Problem Description

bobo has a sequence a1,a2,…,an. He is allowed to swap two adjacent numbers for no more than k times.

Find the minimum number of inversions after his swaps.

Note: The number of inversions is the number of pair (i,j) where 1≤iaj.

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,k (1≤n≤105,0≤k≤109). The second line contains n integers a1,a2,…,an (0≤ai≤109).

Output

For each tests:

A single integer denotes the minimum number of inversions.

#include 

using namespace std;
long long int merge(int *a,int start,int mid,int end,int *t){
    int i=start,j=mid+1,k=start;
    long long int cnt=0;
    while (i<=mid&&j<=end)
    {
        if(a[i]>a[j]){
            t[k++]=a[j++];cnt+=mid-i+1;
        }
        else{
            t[k++]=a[i++];
        }
    }
    while (i<=mid)
    {
        t[k++]=a[i++];
    }
    while (j<=end)
    {
        t[k++]=a[j++];
    }
    for (int i=start;i<=end;i++)  //这里忘了将临时数组的值还回去，这样每次白排序了
    {
        a[i]=t[i];
    }
    return cnt;
}
long long int  merge_sort(int *a,int start,int end,int *t ){
    int mid;
    long long int cnt=0;
    if(start==end){
        t[start]=a[start];
    }
    else{
        mid=(start+end)/2;
        cnt+=merge_sort(a,start,mid,t);//这里也要加cnt，每个递归的mergeSort的值都要加起来，为防止错误，最好用全局变量
        cnt+=merge_sort(a,mid+1,end,t);
        cnt+=merge(a,start,mid,end,t);
    }
    return cnt;
}


int main()
{
    int n,k;
    long long int cnt;
    while (cin>>n>>k)
    {
        int a[n];
        int t[n];
        for (int i=0;i>a[i];
        }
        cnt=merge_sort(a,0,n-1,t);
        long long int result=((cnt-k)>=0)?(cnt-k):0;
        cout<

注意事项

1.一定要注意数字越界。。。。。int 和long我曹溢出了。。。0≤k≤109，cnt数字很大，longlongint和long，int的区别