题目描述
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.
输入
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.
输出
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.
样例输入
5
9
1
0
5
4
3
1
2
3
0
样例输出
6
0
#include <bits/stdc++.h> #define ll long long #define ull unsigned long long #define met(a, x) memset(a,x,sizeof(a)) #define inf 0x3f3f3f3f #define mp make_pair; using namespace std; const int mod = 1e9 + 7; const int N = 1e6 + 10; const int M = 1e5 + 10; int n,C[2*N]; struct node{ int x,y; }s[2*N]; int lowbit(int x){ return x&(-x); } void add(int x,int y){ for(;x<=n;x+=lowbit(x)) C[x]+=y; } int getsum(int x){ int ans=0; for(;x;x-=lowbit(x)){ ans+=C[x]; } return ans; } bool cmp(node a,node b){ if(a.x==b.x) return a.y>b.y; return a.x>b.x; } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin>>n; for(int i=1;i<=n;i++){ cin>>s[i].x; s[i].y=i; } sort(s+1,s+1+n,cmp); ll ans=0; for(int i=1;i<=n;i++){ add(s[i].y,1); ans+=getsum(s[i].y-1); } cout<<ans<<endl; return 0; }