题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4911
3 1 2 2 1 3 0 2 2 1
1 2
#include<stdio.h>
int is1[112345],is2[112345];// is1为原数组,is2为临时数组,n为个人定义的长度
__int64 merge(int low,int mid,int high)
{
int i=low,j=mid+1,k=low;
__int64 count=0;
while(i<=mid&&j<=high)
if(is1[i]<=is1[j])// 此处为稳定排序的关键,不能用小于
is2[k++]=is1[i++];
else
{
is2[k++]=is1[j++];
count+=j-k;// 每当后段的数组元素提前时,记录提前的距离
}
while(i<=mid)
is2[k++]=is1[i++];
while(j<=high)
is2[k++]=is1[j++];
for(i=low;i<=high;i++)// 写回原数组
is1[i]=is2[i];
return count;
}
__int64 mergeSort(int a,int b)// 下标,例如数组int is[5],全部排序的调用为mergeSort(0,4)
{
if(a<b)
{
int mid=(a+b)/2;
__int64 count=0;
count+=mergeSort(a,mid);
count+=mergeSort(mid+1,b);
count+=merge(a,mid,b);
return count;
}
return 0;
}
int main()
{
int n, x;
__int64 k;
__int64 sum;
while(scanf("%d%I64d",&n,&k)!=EOF)
{
for(int i=0;i<n;i++)
{
scanf("%d",&x);
is1[i] = x;
}
__int64 ans=mergeSort(0,n-1);
sum=0;
printf("%I64d\n",ans-k>0?ans-k:0);
}
return 0;
}hdu 4911 Inversion(求逆序数),布布扣,bubuko.com
原文地址:http://blog.csdn.net/u012860063/article/details/38395953