usaco Chapter 3 section 3.2 Stringsobits

最新推荐文章于 2016-03-21 12:26:00 发布

原创最新推荐文章于 2016-03-21 12:26:00 发布 · 457 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#input #c

编程算法专栏收录该内容

50 篇文章

订阅专栏

本文介绍了一种算法，该算法用于在一个有序的位串集合中查找特定元素。位串长度固定为N，且每个位串包含不超过L个1。通过输入一个索引I，程序能够返回集合中第I个元素的具体值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/*
ID: niepeng1
LANG: C++
TASK:kimbits
*/
/*
Stringsobits
Kim Schrijvers
Consider an ordered set S of strings of N (1 <= N <= 31) bits. Bits, of course, are either 0 or 1.

This set of strings is interesting because it is ordered and contains all possible strings of length N that have L (1 <= L <= N) or fewer bits that are `1'.

Your task is to read a number I (1 <= I <= sizeof(S)) from the input and print the Ith element of the ordered set for N bits with no more than L bits that are `1'.

*/
#include<iostream>
using namespace std;

int main()
{
unsigned long int map[33][33];
unsigned long int N,L,I;
int i,j;
freopen("kimbits.in","r",stdin);
freopen("kimbits.out","w",stdout);
scanf("%lu %lu %lu",&N,&L,&I);
for(i=0;i<=N;i++)
{
  map[0][i]=1;
  map[i][0]=1;
}
for(i=1;i<=N;i++)
  for(j=1;j<=N;j++)
  {
   if(j>i)
    map[i][j]=map[i][i];
   else
    map[i][j]=map[i-1][j]+map[i-1][j-1];
  }
for(i=N-1;i>=0;i--)
{
  if(map[i][L] < I)
  {
   printf("1");
   I-=map[i][L];
   L--;
  }
  else
   printf("0");
}
printf("/n");
return 0;
}