PAT T1010 Lehmer Code

跟1009几乎是同一道题~

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+14;
int a[maxn];
int c[maxn*8];
int r[maxn];
int lowbit (int x) {
    return x&-x;
}
int main () {
    int N;
    scanf ("%d",&N);
    for (int i=0;i<N;i++) {
        scanf ("%d",&a[i]);
        a[i]+=10010;
    }
    for (int i=N-1;i>=0;i--) {
        for (int j=a[i]-1;j>0;j-=lowbit(j)) 
        r[i]+=c[j];
        for (int j=a[i];j<maxn;j+=lowbit(j))
        c[j]++;
    } 
    for (int i=0;i<N;i++) {
        if (i!=0) printf (" ");
        printf ("%d",r[i]);
    }
    return 0;
}

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转载自www.cnblogs.com/zhanglichen/p/12302852.html