数状数组

1.循环右移，老套路了，直接开2n数组

然后利用树状数组进行区间求和和单点修改,每次减去之前以及出现过的值

Problem - E - Codeforces

#include<bits/stdc++.h>
using namespace std;

const int N = 1e6 + 10;

int t;
int c[N * 2];
int sum(int x)
{
    int res = 0;
     while(x){
         res += c[x];
         x -= x & -x;
     }
     return res;
} 
void add(int x,int v,int n)
{
    while(x <= n)
    {
        c[x] += v;
        x += x & -x;
    }
}
struct Node{
    int x,id;
};
vector<Node> b[N * 2];
int a[N];
int ans[N];
int main()
{
    scanf("%d", &t);
    while(t -- )
    {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= 2 * n;i ++) c[i] = 0,b[i].clear();
        for(int i = 1; i <= n; i ++){ //对右端点进行排序 
            scanf("%d", &a[i]);
            if(i <= a[i]){ //b数组的存的右端点的值是唯一的
                b[a[i]].push_back({i,a[i]}); //右边a[i] 表示询问的编号
                b[a[i] + n].push_back({i + n,a[i]});
            }else {
                b[a[i] + n].push_back({i,a[i]}); //i表示左端点，a[i] + n 表示右端点
            }
        }
        for(int q = 1; q <= 2 * n; q ++){
            for(auto [x, id] : b[q]){
                if(x <= n) ans[id] = q - x - (sum(q - 1) - sum(x));
                add(x,1,2 * n);
            }
        }
        for(int i = 1; i <= n; i ++) printf("%d ",ans[i]);
        cout << endl;
    }
    return 0;
}