Stringsobits（模拟）

描述

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'.

输入

A single line with three space separated integers: N, L, and I.

输出

A single line containing the integer that represents the Ith element from the order set, as described.

样例输入

5 3 19

样例输出

 10011

题目大意：

求一个长度为N的有L个1的第I大的二进制数。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int dp[35][35],ans[35];
void f(int n,int l,ll m)
{
    ll s=0,la;
    for(int i=0;i<=n;i++)
    {
        la=s;s=0;
        for(int j=0;j<=l;j++)
        {
            s+=dp[i][j];
            if(s>=m)
            {
                ans[i]=1;
                return f(n-1,l-1,m-la);
            }
        }
    }
}
int main()
{
    int n,l;
    ll m;
    scanf("%d%d%I64d",&n,&l,&m);
    for(int i=0;i<=n;i++)
        dp[i][0]=1;
    for(int i=1;i<=n;i++)///i位有j位为1的方案数等价于C（i,j）
        for(int j=1;j<=i;j++)
            dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
    f(n,l,m);
    for(int i=n;i>=1;i--)
        printf("%d",ans[i]);
    return 0;
}