nyoj 117 求逆序数

#include<stdio.h>
int A[1000001],T[1000001];long long count=0;
void merge_s(int *A,int x,int y,int *T)
{
if(x==y) return ;
int m=(x+y)>>1;
merge_s(A,x,m,T);
merge_s(A,m+1,y,T);
int i=x,j=m+1,pos=x;
while(i<=m&&j<=y){
if(A[i]<=A[j]) T[pos++]=A[i++];
else {T[pos++]=A[j++];count+=m-i+1;}
}
while(i<=m)
T[pos++]=A[i++];
while(j<=y) T[pos++]=A[j++];
for(i=x;i<=y;i++)
A[i]=T[i];
}
int main()
{
int t,N,i;scanf("%d",&t);
while(t--)
{
count=0;
scanf("%d",&N);
for(i=0;i<N;i++)
scanf("%d",&A[i]);
merge_s(A,0,N-1,T);
printf("%lld\n",count);
}
return 0;
}

#include <iostream>
#include <cstring>
#include <cstdio>
#include
using namespace std;
const int N = 1000005;
int reflect
,c
,n;
struct node
{
int val;
int pos;
}s
;
bool cmp(node
4000
x,node y)
{
if(x.val == y.val) return x.pos < y.pos;
return x.val < y.val;
}
int lowbit(int x)
{
return x&(-x);
}
void update(int x)
{
while(x <= n)
{
c[x] += 1;
x += lowbit(x);
}
}
long long getsum(int x)//统计比x小的数
{
long long sum = 0;
while(x > 0)
{
sum += c[x];
x -= lowbit(x);
}
return sum;
}
int main()
{
int T,i;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i = 1; i <= n; i ++)
scanf("%d",&s[i].val), s[i].pos = i;
sort(s+1,s+n+1,cmp);
for(i = 1; i <= n; i ++)
reflect[s[i].pos] = i;
memset(c,0,sizeof(c));
long long ans = 0;
for(i = 1; i <= n; i ++)
{
update(reflect[i]);
ans += i - getsum(reflect[i]);//i表示已经插入的数的个数
}
printf("%lld\n",ans);
}
return 0;
}