UVaLive 11525 Permutation (线段树)

题意：有一个由1到k组成的序列，最小是1 2 … k，最大是 k k-1 … 1，给出n的计算方式，n = s0 * (k - 1)! + s1 * (k - 2)! +… + sk-1 * 0!，

给出s1…sk，输出序列里第n大的序列。

析：我们先看第一数，如果第一个数是2，那么它前面至少有(k-1)!个排列，然后1开头肯定比2要小，同理，再考虑第二个数，在考虑再二数时，

要注意把已经搞定的数去掉，所以我们用线段树进行单点更新，当然也可以用二分+数状数组。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int sum[maxn<<2];
 
void push_up(int rt){  sum[rt] = sum[rt<<1] + sum[rt<<1|1];  }
 
void build(int l, int r, int rt){
  if(l == r){
    sum[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){
    sum[rt] = 0;  return l;
  }
  int m = l+r >> 1;
  int ans;
  if(sum[rt<<1] >= x)  ans = query(x, lson);
  else  ans = query(x-sum[rt<<1], rson);
  push_up(rt);
  return ans;
}
 
int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    build(1, n, 1);
    for(int i = 0; i < n; ++i){
      if(i)  putchar(' ');
      scanf("%d", &m);
      printf("%d", query(m + 1, 1, n, 1));
    }
    printf("\n");
  }
  return 0;
}