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'.
PROGRAM NAME: kimbits
INPUT FORMAT
A single line with three space separated integers: N, L, and I.SAMPLE INPUT (file kimbits.in)
5 3 19
OUTPUT FORMAT
A single line containing the integer that represents the Ith element from the order set, as described.SAMPLE OUTPUT (file kimbits.out)
10011
用组合数缩小范围
比如对于样例中的10000，小于它的有C43+C42+C41+C04个,（右边4个0，选几个变成1）
从左往右一个一个确定是不是1（根据范围，若比它小的数的个数小于题目所求，则这一位为1）
注意取一个1后，L--
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define name "kimbits"
using namespace std;
int C[32][32];
long long n,l,x;
int get(int a,int b)
{
int ret=0;
for (int i=0;i<=b;i++)
ret+=C[a][i];
return ret;
}
int main()
{
freopen(name ".in","r",stdin);
freopen(name ".out","w",stdout);
cin>>n>>l>>x;
int i,j,tmp;
C[1][0]=C[0][0]=C[1][1]=1;
for (i=2;i<=n;i++)
for (j=0;j<=i;j++)
C[i][j]=C[i-1][j]+C[i-1][j-1];
if (x==1) {for (i=1;i<=n;i++)cout<<"0"; cout<<endl;return 0;}
for (i=1;i<=n;i++)
{
tmp=get(i-1,l);
if (tmp>=x) {i--;break;}
}
if (i==n+1) i=n;
tmp=get(i-1,l);
l--;
for (j=i+1;j<=n;j++) cout<<"0";
cout<<"1";
for (j=i-1;j>=1;j--)
{
int g=get(j-1,l);
if (tmp+g>=x) cout<<"0";
else {cout<<"1";l--;tmp+=g;}
}
cout<<endl;
return 0;
}