Description
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.
Input
The input contains several test cases. Every test case begins with a line that contains a single integer n <= 100,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
Output
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
Sample Input
5
9
1
0
5
4
3
1
2
3
0
Sample Output
6
0
Notice
The answer may be very large. You can use "long long" to store the answer and "%lld" to print it.
用歸併排序做很簡單。當然也可以用樹狀數組來做。
我是用的歸併
#include <stdio.h>
#include <string.h>
int N, A[500010], T[500010];
long long ans;
void Merg_Sort(int x,int y)
{
if(y-x<=1) return;
int mid;
mid = x + (y-x)/2;
Merg_Sort(x,mid);
Merg_Sort(mid,y);
int p = x, q = mid, i=x;
while(p<mid || q<y)
{
if(q>=y || (p<mid && A[p] <= A[q])) T[i++] = A[p++];
else//else的條件是(p==mid || A[q] < A[p])
{
if(p<mid)
ans+=(mid-p);//由於是p<mid,所以此時也就是相當於 A[q] < A[p]
T[i++] = A[q++]; //上面同時A[p]是第一個<A[q]的數，所以+後面還有的數(mid-p)
}
}
for(i=x;i<y;i++)
{
A[i] = T[i];
}
}
int main()
{
while(scanf("%d", &N) && N)
{
int i;
memset(A,0,sizeof(A));
memset(T,0,sizeof(T));
for(i=0;i<N;i++)
{
scanf("%d", &A[i]);
}
ans = 0;
Merg_Sort(0,N);
printf("%lld\n",ans);//結果會超int32
}
return 0;
}