Gym - 102082G
https://vjudge.net/problem/2198225/origin
对于数列中任意一个数，要么从最左边到它不递减，要么从最右边到到它不递减，为了满足这个条件，就要移动，而移动的最少步数就是逆序对数。所以这个数要么往左移动，要么往右移动，所以两个取最小就好了
#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cmath> #include<ctime> #include<set> #include<map> #include<stack> #include<cstring> #define inf 2147483647 #define ls rt<<1 #define rs rt<<1|1 #define lson ls,nl,mid,l,r #define rson rs,mid+1,nr,l,r #define N 1000010 #define For(i,a,b) for(long long i=a;i<=b;i++) #define p(a) putchar(a) #define g() getchar()using namespace std; long long n,len,ans,k; long long a[N],b[N],c[N],temp,d1[N],d2[N]; long long t[N]; void in(long long &x){long long y=1;char c=g();x=0;while(c<'0'||c>'9'){if(c=='-')y=-1;c=g();}while(c<='9'&&c>='0'){x=(x<<1)+(x<<3)+c-'0';c=g();}x*=y; } void o(long long x){if(x<0){p('-');x=-x;}if(x>9)o(x/10);p(x%10+'0'); }void modify(long long k){for(;k<=n;k+=k&(-k))t[k]++; }long long getnum(long long k){long long cnt=0;for(;k>0;k-=k&(-k))cnt+=t[k];return cnt; }int main(){in(n);For(i,1,n){in(a[i]);b[i]=a[i];c[i]=a[i];}sort(b+1,b+n+1);sort(c+1,c+n+1);len=unique(b+1,b+n+1)-b-1;For(i,1,n)a[i]=lower_bound(b+1,b+n+1,a[i])-b;For(i,1,n){modify(a[i]);d1[i]=i-getnum(a[i]);}memset(t,0,sizeof(t));for(long long i=n;i;i--){modify(a[i]);d2[i]=n-i-getnum(a[i])+1;}For(i,1,n)ans+=min(d1[i],d2[i]);o(ans);return 0; }
