hust 1433（树状数组）

http://acm.hust.edu.cn/thx/problem.php?id=1433
方法一：
用两遍树状数组，分别求出a[i]与前面的数字组成的逆序对数记为c[i]，
和与后面的数字组成的逆序对数记为d[i]，以及总的逆序对数记为tot，
则
#include <cstdio>
#include <iostream>
const long long max_n = 100010;
long long a[max_n], c[max_n];
long long b[max_n];

long long lowbit(long long x) 
{
    return x & -x;
}

void insert(long long x, long long n) 
{
    for (; x <= n; x += lowbit(x))
        c[x]++;
}

long long sum(long long x) 
{
    long long s = 0;
    for (; x; x -= lowbit(x)) 
        s += c[x];
    return s;
}

void solve(long long n)
{
    for (long long i = 1; i <= n; i++) 
        scanf("%lld", &a [i]);

    memset(c, 0, sizeof (c));
    b[n + 1] = 0;

    for (long long i = n; i ; i--)
    {
        b[i] = sum(a[i]) + b[i + 1];
        insert(a[i], n);
    }
    long long total = b[1], before = 0;
    memset(c, 0, sizeof (c));

    for (long long i = 1; i < n; i++)
    {
        before += i - 1 - sum(a[i]);
        insert(a[i], n);
        b[i] = total - before - b[i + 1];
    }

    for (int i = 1; i < n - 1; i++) 
        printf("%lld ", b[i]);
    printf("%lld\n", b[n - 1]);
}

int main()
{
    long long n;
    while (scanf("%lld", &n) != EOF)
    {
        solve(n);
    }
}


方法二：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define  max 500001
long long  tree[max];
long long x[max];
long long b[max];
long long  getsum(long long idx)
{
    long long  sum=0;
    while(idx>0)
    {
        sum+=tree[idx];
        idx-=(idx&-idx);
    }
    return sum;
}
void add(long long idx)
{
    while(idx<=max)
    {
        tree[idx]+=1;
        idx+=(idx&-idx);
    }
}
int main()
{
    long long  n,i;
    while(scanf("%lld",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&x[i]);
            b[i]=b[i-1]-(i-1-getsum(x[i]));
            add(x[i]);
            b[i]+=x[i]-getsum(x[i]);
        
        }
        for(i=1;i<n;i++) 
        {
            if(i==1)    printf("%lld",b[i]);
            else printf(" %lld" ,b[i]);
        }
        printf("\n");
    }
}


方法三：


#include<stdio.h>
#define MAXN 100010
long long  a[MAXN],b[MAXN];
int main()
{
  long long  i,n;
  while(scanf("%lld",&n)!=EOF)
  {
      b[0]=0;
    for(i=1;i<=n;i++)  scanf("%lld",&a[i]);
    for(i=1;i<n;i++)    b[i]=b[i-1]+a[i]-i;
    for(i=1;i<n-1;i++) printf("%lld ",b[i]);
    printf("%lld\n",b[i]);

  }
  return 0;
}