Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 24310 Accepted: 8696
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 < 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
Source
Waterloo local 2005.02.05
#include <stdio.h> #include <stdlib.h> #include <string.h> #define MAXN 500010 int n, a[MAXN], c[MAXN]; long long MergeSort(int left, int right){ int i, j, k, mid; long long ans; if (left >= right) return 0; mid = (left + right) / 2; ans = MergeSort(left, mid) + MergeSort(mid + 1, right); for (i = left, j = mid + 1, k = left; i <= mid && j <= right; k++){ if (a[i] < a[j]) c[k] = a[i++]; else{ c[k] = a[j++]; ans += mid - i + 1; } } for (; i <= mid; i++, k++) c[k] = a[i]; for (; j <= right; j++, k++) c[k] = a[j]; for (i = left; i <= right; i++) a[i] = c[i]; /* printf("m: (%d, %d) = %d\n", left, right, ans); for (i = left; i <= right; i++){ printf("%d ", a[i]); } printf("\n"); */ return ans; } int main(){ int i; while(scanf("%d", &n), n != 0){ for (i = 0; i < n; i++) scanf("%d", &a[i]); printf("%lld\n", MergeSort(0, n - 1)); } return 0; } /* 求逆序数,可以归并排序,离散化+树状数组 这里是归并排序 375MS 在g++里排整100... WA了一次,结果要long long...最坏情况逆序排列,需交换n*(n+1)/2 */