题目大意：给你N个数，每次只能交换相邻的两个数，问至少需要交换几次，才能使数变成升序

解题思路：需要交换的次数就是逆序对的个数，每交换一次，就相当于减少了一次逆序，而最后的升序，刚好逆序对是为0的，所以有多少的逆序对就要交换几次

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 500010;
int n;
int val[N], val2[N], bit[N];

void init() {
    for (int i = 1; i <= n; i++) {
        scanf("%d", &val[i]);
        val2[i] = val[i];
    }
    sort(val + 1, val + 1 + n);
    memset(bit, 0, sizeof(bit));
}

int find(int num) {
    int l = 1, r = n;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (val[mid] == num) return mid;
        else if (val[mid] > num) r = mid - 1;
        else l = mid + 1;
    }
    return -1;
}

inline int lowbit(int x) {
    return x & (-x);
}

int Query(int x) {
    int ans = 0;
    while (x) {
        ans += bit[x];
        x -= lowbit(x);
    }
    return ans;
}

void Modify(int x, int w) {
    while (x < N) {
        bit[x] += w;
        x += lowbit(x);
    }
}

void solve() {
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
        int pos = find(val2[i]);
        Modify(pos, 1);
        ans += i - Query(pos);
    }
    printf("%lld\n", ans);
}

int main() {
    while (scanf("%d", &n) != EOF && n ) {
        init();
        solve();
    }
    return 0;
}