解题报告
题意:
求逆序数。
思路:
线段树离散化处理。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#define LL long long
using namespace std;
LL sum[2001000],num[501000],_hash[501000];
void push_up(int rt)
{
sum[rt]=sum[rt*2]+sum[rt*2+1];
}
void update(int rt,int l,int r,int p,LL v)
{
if(l==r)
{
sum[rt]=+v;
return;
}
int mid=(l+r)/2;
if(p<=mid)update(rt*2,l,mid,p,v);
else update(rt*2+1,mid+1,r,p,v);
push_up(rt);
}
LL q_sum(int rt,int l,int r,int ql,int qr)
{
if(ql>r||qr<l)return 0;
if(ql<=l&&r<=qr)return sum[rt];
int mid=(l+r)/2;
return q_sum(rt*2,l,mid,ql,qr)+q_sum(rt*2+1,mid+1,r,ql,qr);
}
int main()
{
int n,i,j;
while(~scanf("%d",&n))
{
LL ans=0;
if(!n)break;
memset(_hash,0,sizeof(_hash));
memset(num,0,sizeof(num));
memset(sum,0,sizeof(sum));
for(i=0; i<n; i++)
{
scanf("%LLd",&num[i]);
_hash[i]=num[i];
}
sort(_hash,_hash+n);
int m=unique(_hash,_hash+n)-_hash;
for(i=0; i<n; i++)
{
int t=lower_bound(_hash,_hash+m,num[i])-_hash+1;
ans+=q_sum(1,1,m,t+1,m);
update(1,1,m,t,1);
}
printf("%lld\n",ans);
}
return 0;
}
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 41278 | Accepted: 14952 |
Description

Ultra-QuickSort produces the output
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