# 【逆序对】 模板

#include <cstdio>
#define ll long long
using namespace std;
const int maxn = 500001;
ll a[maxn], s[maxn], ans = 0, n;//ans用来记录逆序对数量
void merge_sort(ll l,ll r)
{
if(l == r) return ;
ll mid = (l + r)>>1;
merge_sort(l,mid);
merge_sort(mid+1,r);//不断划分成两个数列

ll i = l, j = mid+1, k =l;

while(i <= mid&&j <= r)
{
if(a[i]<=a[j])
{
s[k] = a[i],k++,i++;
}

else
{
s[k] = a[j],k++,j++;
ans+=mid-i+1;
}
}
while(i <= mid)
s[k] = a[i],k++,i++;
while(j<=r)
s[k] = a[j],k++,j++;
for(int i = l; i <= r; i++)
a[i] = s[i];
}
int main()
{
scanf("%lld",&n);
for(int i = 1; i <= n; i++)
{
scanf("%lld",&a[i]);
}
merge_sort(1,n);
/*for(int i = 1; i <= n; i++)
{
printf("%d ",a[i]);
}*/
printf("%lld",ans);
return 0;
}


BIT：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
const int maxn = 1e6 + 10;
ll n, m, A[maxn], B[maxn], ans, L[maxn];
class BIT{
public:
ll tree[maxn];
void update(ll pos, ll val)
{
while(pos <= n)
{
tree[pos] += val;
pos += lowbit(pos);
}
}
ll query(ll pos)
{
ll res = 0;
while(pos)
{
res += tree[pos];
pos -= lowbit(pos);
}
return res;
}
private:
ll lowbit(ll x)
{
return x & -x;
}
}T;
int Search(ll x)
{
return lower_bound(B + 1, B + 1 + m, x) - B;
}
int main()
{
scanf("%lld",&n);
for(int i = 1; i <= n; i++) scanf("%lld",&A[i]), B[i] = A[i];
sort(B + 1, B + 1 + n);
m = unique(B + 1, B + 1 + n) - B - 1;
for(int i = 1; i <= n; i++) A[i] = Search(A[i]);
for(int i = 1; i <= n; i++)
{
T.update(A[i], 1);
L[i] = i - T.query(A[i]);
}
for(int i = 1; i <= n; i++) ans += L[i];
printf("%lld",ans);
return 0;
}


posted @ 2019-06-13 21:26  Misaka_Azusa  阅读(164)  评论(0编辑  收藏
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