題意：一個逆序對罰錢x元，現在給你交換的機會，每交換任意相鄰兩個數花錢y，問你最少付多少錢

思路：最近在補之前還沒過的題，發現了這道多校的題。顯然，交換相鄰兩個數逆序對必然會變化+1或者-1，那我們肯定是-1操作。那麼顯然問題就變成了求逆序對數*min（x，y）。樹狀陣列求逆序對數。

程式碼：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include
 
<cstring>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 100000 + 10;
const int seed = 131;
const ll MOD = 100000007;
const int INF = 0x3f3f3f3f;
int a[maxn], b[maxn], node[maxn];
int lowbit(int x){
    return x&(-x);
}
void update(int
 
 x){
    for(int i = x; i < maxn; i += lowbit(i))
        node[i]++;
}
ll query(int x){
    ll ret = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        ret += node[i];
    return ret;
}
int main(){
    int n, x, y;
    while(~scanf("%d%d%d", &n, &x, &y)){
        ll ans 
 
= 0;
        memset(node, 0, sizeof(node));
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]), b[i] = a[i];
        sort(b + 1, b + n + 1);
        for(int i = 1; i <= n; i++)
            a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;
        for(int i = n; i >= 1; i--){
            ans += query(a[i] - 1);
            update(a[i]);
        }
        printf("%lld\n", min(x, y) * ans);
    }
    return 0;
}