POJ2299Ultra-QuickSort（64位long long）

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 31628 Accepted: 11281

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

59105431230

Sample Output

60

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 500010;long long cnt = 0;int a[maxn], t[maxn];void merge_sort(int *A, int x, int y, int* T) {//contain [x,y]    if(x==y) return ;   // printf("x = %d, y = %d\n", x, y);    if(y-x>=1) {        int m = (x+y)/2;        int p = x, q = m+1, i = x;        merge_sort(A, x, m, T);        merge_sort(A, m+1, y, T);        while(p <= m || q <= y) {            if(p<=m && q<=y && A[p]<=A[q]) {                T[i++] = A[p++];            }            if(p<=m && q<=y && A[p]>A[q]) {                cnt =  cnt + (m-p+1);             //   printf("cnt = %d\n", cnt);                T[i++] = A[q++];            }            while(q>y && p<=m) {                T[i++] = A[p++];            }            while(p>m && q<=y) {                T[i++] = A[q++];            }        }        for( i = x; i <= y; i++) {            A[i] = T[i];        }    }}int main(){    int n;    while((scanf("%d", &n) != EOF) && n) {        for(int i = 0; i < n; i++) {            scanf("%d", &a[i]);        }        merge_sort(a, 0, n-1, t);        printf("%lld\n", cnt);        cnt = 0;    }    return 0;}/**********************59 1 0 5 489 1 0 5 4 3 8 7**************/