传送门：点击打开链接
题意：告诉你每个数字的逆序数，求原序列
思路：从1开始考虑，利用线段树记录空白位置的数量，然后利用线段树维护插入位置经行插入
#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck printf("fuck")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;const int MX = 10000 + 5;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,n,1int S[MX << 2], A[MX], B[MX];void push_up(int rt) {S[rt] = S[rt << 1] + S[rt << 1 | 1];
}void build(int l, int r, int rt) {if(l == r) {S[rt] = 1;return;}int m = (l + r) >> 1;build(lson);build(rson);push_up(rt);
}int query(int x, int l, int r, int rt) {if(l == r) {S[rt]--;return l;}int m = (l + r) >> 1, ret;if(x <= S[rt << 1]) ret = query(x, lson);else ret = query(x - S[rt << 1], rson);push_up(rt);return ret;
}int main() {int n; //FIN;while(~scanf("%d", &n)) {build(root);for(int i = 1; i <= n; i++) {scanf("%d", &A[i]);}bool sign = false;for(int i = 1; i <= n; i++) {if(S[1] < A[i] + 1) {sign = true;break;}int p = query(A[i] + 1, root);B[p] = i;}if(sign) printf("No solution\n");else for(int i = 1; i <= n; i++) {printf("%d%c", B[i], i == n ? '\n' : ' ');}}return 0;
}