UVaLive 11525 Permutation (线段树)

时间：2023-01-30 10:49:29

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;

}

　　

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