USACO3.2.2--Stringsobits

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
题解：看到这个果断枚举了。。。速度码出来了。。不过TLE了。。。N=31,枚举量等于2^31。。。线性的枚举也得超时啊。。。这样的话只能把第I个符合要求的数构造出来。
我们分两步：
第一步：计算长度为i,1的个数不超过j的01字符串组成的集合的个数，用f[i][j]表示。
那么DP方程为f[i][j]=f[i-1][j]+f[i-1][j-1]（分别表示开头为‘0’和开头为‘1’的情况）。
边界条件：f[i,0]=1,f[0,j]=1。
第二步：这部就是利用上述DP方程构造出第I个符合要求的01字符串。估计我说得很渣。。。直接引用nocow上的话好了：设所求串为S,假设S的位中最高位的1在自右向左第K+1位，那么必然满足F[K,L]< I,F[K+1,L] >=I,这样的K是唯一的。所以S的第一个1在从右至左第K+1位.因为有F[K,L]个串第K+1位上为0，所以所求的第I个数的后K位就应该是满足"位数为K且串中1不超过L-1个"这个条件的第I-F[K,L]个数。

View Code

/*

ID:spcjv51

PROG:kimbits

LANG:C

*/

#include<stdio.h>

#include<string.h>

int main(void)

{

    freopen("kimbits.in","r",stdin);

    freopen("kimbits.out","w",stdout);

    long i,j,n,l;

    long long m;

    long f[32][32];

    scanf("%ld%ld%lld",&n,&l,&m);

    memset(f,0,sizeof(f));

    for(i=0;i<=n;i++)

    {

        f[0][i]=1;

        f[i][0]=1;

    }

    for(i=1;i<=n;i++)

    for(j=1;j<=l;j++)

    f[i][j]=f[i-1][j]+f[i-1][j-1];

    for(i=n;i>=1;i--)

    if(f[i-1][l]<m)

    {

        printf("1");

        m-=f[i-1][l];

        l--;

    }

    else

    printf("0");

    printf("\n");

    return 0;

}