题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5775

Bubble Sort

for(int i=1;i<=N;++i)
    for(int j=N,t;j>i;—j)
        if(P[j-1] > P[j])
            t=P[j],P[j]=P[j-1],P[j-1]=t;

2
3
3 1 2
3
1 2 3

Case #1: 1 1 2
Case #2: 0 0 0

Hint
In first case, (3, 1, 2) -> (3, 1, 2) -> (1, 3, 2) -> (1, 2, 3)
the leftmost place and rightmost place of 1 is 1 and 2, 2 is 2 and 3, 3 is 1 and 3
In second case, the array has already in increasing order. So the answer of every number is 0.
 

解题思路：

我们从小到大考虑每一个数组，计算出他左边比他小的个数，再用n减去就是右边比他小的数量了，采用树状数组数组的方式来进行运算。

详见代码。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

#define N 100000+10
int c[N],a[N],ans[N];

int lowbit(int k)
{
    return k&((k)xor(k-1));
}

void add(int num,int k)
{
    while (k<=N)
    {
        c[k]+=num;
        k+=lowbit(k);
    }
}

int sum(int k)
{
    int s=0;
    while (k)
    {
        s+=c[k];
        k-=lowbit(k);
    }
    return s;
}


int main()
{
    int t,s,Case=1;
    scanf("%d",&t);
    while (t--)
    {
        int n,k,Max,Min;
        scanf("%d",&n);
        memset(c,0,sizeof(c));
        for (int i=1; i<=n; i++)
        {
            k=0;
            scanf("%d",&a[i]);
            add(1,a[i]);//在a[i]的位置加1
            s=a[i]-sum(a[i]);//a[i]后面有多少个比当前数值小的
            Max=max(a[i],i+s);
            Min=min(a[i],i);
            ans[a[i]]=Max-Min;
        }
        printf ("Case #%d:",Case++);
        for (int i=1; i<=n; i++)
            printf (" %d",ans[i]);
        printf ("\n");
    }
    return 0;
}

hdu 5775 Bubble Sort（2016 Multi-University Training Contest 4——树状数组）

原文：http://blog.csdn.net/qiqi_skystar/article/details/52068553