SPOJ INVCNT - Inversion Count

题目链接：http://www.spoj.com/problems/INVCNT/

题目大意：求数组中逆序数的个数。即满足 i < j, a[i] > a[j] 的数的个数。

解题思路：可以使用归并排序求逆序数，每次归并的时候记录一下：

假如归并 va = {3 5 6} 和 vb = {1 2 3}， 那么 对于3来说，逆序数有2个；对于5，6来说有3个。简单记录一下即可。

 

也可以使用树状数组，枚举每个数字查询比他大的数字出现了几个。不过当数组元素过大的时候需要离散化。

代码：

const int maxn = 1e6 + 5;
int n;
int a[maxn];
ll ans = 0;

VI mymerge(VI va, VI vb){
    VI vc;
    int i = 0, j = 0;
    while(i < va.size() || j < vb.size()){
        while((i >= va.size() && j < vb.size()) || (j < vb.size() && va[i] > vb[j])){
            vc.push_back(vb[j++]);
        }
        if(i < va.size()) {
            vc.push_back(va[i++]);
            ans += j;
        }
    }
    return vc;
}
VI mergesort(VI va){
    int t = va.size();
    if(t > 1){
        VI vb, vc;
        for(int i = 0; i < t / 2; i++) vb.push_back(va[i]);
        for(int i = t / 2; i < va.size(); i++) vc.push_back(va[i]);
        vb = mergesort(vb); 
        vc = mergesort(vc);
        va = mymerge(vb, vc);
    }
    return va;
}
void solve(){
    ans = 0;
    VI va;
    for(int i = 0; i < n; i++) va.push_back(a[i]);
    mergesort(va);
    printf("%lld\n", ans);
    va.clear();
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        solve();
    }
}

题目：

INVCNT - Inversion Count

#graph-theory #number-theory #shortest-path #sorting #bitmasks

 

Let A[0...n - 1] be an array of n distinct positive integers. If i < j and A[i] > A[j] then the pair (i, j) is called an inversion of A. Given n and an array A your task is to find the number of inversions of A.

Input

The first line contains t, the number of testcases followed by a blank space. Each of the t tests start with a number n (n <= 200000). Then n + 1 lines follow. In the ith line a number A[i - 1] is given (A[i - 1] <= 10^7). The (n + 1)th line is a blank space.

Output

For every test output one line giving the number of inversions of A.

Example

Input:
2

3
3
1
2

5
2
3
8
6
1


Output:
2
5