spoj的时间限制超出错误。我如何改善解决方案？

Question

We are given an array a[1..N] . For each element a[i] in the array, we note down the sum of all the elements which are smaller and occur before the present element. I need to calculate the total sum for every element in the array.

Constraints:

1<=N<=10^5

All elements will be between 0 and 10^6.

Here is the link to my question: http://www.spoj.com/problems/DCEPC206/ . I'm using the approach shown below, but I'm getting TIME LIMIT EXCEEDED error on SPOJ. How can I improve my solution?

include 

int main()

{

long n,a[100000],i,j,sum;

printf("enter the number of elements");

  scanf("%ld",&n);

printf("enter the elements of the array");

for(i=0;i<n;i++)

      scanf("%ld",&a[i]);

sum=0;


for(i=1;i<n;i++)

     for(j=i-1;j>=0;j--)

         if(a[i]>a[j])

              sum+=a[j];

printf("\n%ld",sum);

return 0;

}

Answer 1

Yours is a trivial implementation which takes time of the order of O(n^2) but for the solution to get accepted you have to use Divide and Conquer as you do in Merge Sort which takes only O(NLogN) time compared to simpler sorting algorithms like Bubble Sort,etc.

For this you simply can change the Mergesort implementation a little bit simply by adding 2-3 lines of code in it. For understanding this better go through question of counting inversions in an array.( http://www.geeksforgeeks.org/counting-inversions/ ) Then you will realize you simply have to consider pairs opposite in nature to inversion and add all the smaller elements of all such pairs . For example - in an array 1,4,2,5 consider 4,2 is inversion but we have to consider pairs like 2,5 and 1,2 to get the solution. In every such pair keep adding the left no. ( Think hard about how it is doing our job !! )

For your reference go thoroughly through this merge sort code in which i have small changes to get a correct accepted solution.(sum variable stores the resultant value)

#include <stdio.h>
#include <stdlib.h>
long long int sum;
void merge(long long int c[],long long int arr[],long long int start,long long int middle,long long int end)
{
    long long int i=0,j=start,k=middle;
    while((j<middle)&&(k<end))
    {
        if(arr[j]<arr[k])
        {
            sum=sum+((end-k)*arr[j]);
            c[i]=arr[j];
            i++;j++;
        }
        else
        {
            c[i]=arr[k];
            i++;k++;
        }
    }
    while(j<middle)
    {
        c[i]=arr[j];
        i++;
        j++;
    }
    while(k<end)
    {
        c[i]=arr[k];
        i++;
        k++;
    }

}


void msort(long long int arr[],long long int start,long long int end)
{
    long long int middle=(start+end)/2;
    if((end-start)==1)
    {   return ;
    }
    msort(arr,start,middle);
    msort(arr,middle,end);
    long long int *c;
    c=(long long int*)malloc(sizeof(long long int)*(end-start));
    merge(c,arr,start,middle,end);
    long long int i,j=0;
    for(i=start;i<end;i++)
    {
        arr[i]=c[j];
        j++;
    }
}

void swap (long long int x[],long long int m,long long int n)
{
  long long int t= x[m];
  x[m]=x[n];
  x[n]=t;
}

int main()
{
    int t,i;
    long long int n,*arr,j;
    scanf("%d",&t);
    for(i=0;i<t;i++)
    {
        scanf("%lld",&n);
        arr = ( long long int * ) malloc ( sizeof(long long int) * n + 10);
        for(j=0;j<n;j++)
        {
            scanf("%lld",&arr[j]);

        }

        sum=0;  
        msort(arr,0,n);
//      for(j=0;j<n;j++)
//          printf("%lld ",arr[j]);
        printf("%lld\n",sum);
    }

    return 0;
}