首先计算出组合数。用cmb_num[i][j]表示i位数中，"1的位数小于等于j"的数的个数。
这样，我们从最左边开始，如果cmb_num[i-1][j]的数大于n，说明第一位为0，因为用i-1位数中"1的位数小于等于j"的数已经大于n个了。
如果小于n，说明第一位为1，需要i位，才能使"1的位数小于等于j"的数大于n个了。既然第一位已经是1了，接下来的i-1位组成的数的1的个数只能小于等于n-1位了。迭代输出每一位即可。
只是要注意溢出的问题以及cmb_num[0][1]。
#include <iostream>
#include <fstream>
using namespace std;
ifstream fin("kimbits.in");
ofstream fout("kimbits.out");
#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif
unsigned int cmb_num[32][32];
void build_cmb_num()
{
for(int i=0;i<32;++i)
cmb_num[i][0] = 1;
for(int i=1;i<32;++i)
for(int j=1;j<=i;++j)
cmb_num[i][j] = cmb_num[i-1][j-1]+cmb_num[i-1][j];
for(int i=0;i<32;++i)
for(int j=1;j<32;++j){
cmb_num[i][j]+=cmb_num[i][j-1];
}
}
void solve()
{
build_cmb_num();
unsigned n,l,i;
in>>n>>l>>i;
for(unsigned idx=n;idx>0;--idx){
if( i> cmb_num[idx-1][l] ){
out<<1;
i-=cmb_num[idx-1][l];
l--;
}else{
out<<0;
}
}
out<<endl;
}
int main(int argc,char *argv[])
{
solve();
return 0;
}
附题：
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'.
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