Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 39529 | Accepted: 14250 |
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.
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 < 500,000 -- 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
注意结果要用long long存储。
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 500002
using std::sort;
int tree[maxn], ori[maxn], hash[maxn];
long long ans;
int getHash(int val, int n)
{
int left = 0, right = n - 1, mid;
while(left <= right){
mid = (left + right) >> 1;
if(val < hash[mid]) right = mid - 1;
else if(val > hash[mid]) left = mid + 1;
else return mid + 1;
}
}
int lowBit(int pos){ return pos & (-pos); }
int getSum(int pos)
{
int sum = 0;
while(pos > 0){
sum += tree[pos];
pos -= lowBit(pos);
}
return sum;
}
void update(int pos, int n)
{
ans += (pos - 1 - getSum(pos - 1));
while(pos <= n){
++tree[pos];
pos += lowBit(pos);
}
}
int main()
{
int n, i;
while(scanf("%d", &n), n){
for(i = 0; i < n; ++i){
scanf("%d", ori + i);
hash[i] = ori[i];
}
sort(hash, hash + n);
for(i = 0; i < n; ++i)
ori[i] = getHash(ori[i], n);
memset(tree, 0, sizeof(tree));
ans = 0;
for(i = 0; i < n; ++i) update(ori[i], n);
printf("%lld\n", ans);
}
return 0;
}
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