USACO: Stringsobits

位串生成算法

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本文介绍了一种用于生成特定条件下的二进制字符串集合的算法。该算法能够根据给定的长度N和最大1的数量L，生成所有可能的有序二进制字符串，并能够找到指定索引I对应的字符串。

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)

解题思路：

1、用一个二维数组cnt[i][j]来记录长度为i的，包含不多于j个1的二进制串的个数，这一步可以这样实现：当j = 0，cnt[i][j] = 1；当j > i，cnt[i][j] = cnt[i][i]；其他情况cnt[i][j] = cnt[i - 1][j] + cnt[i - 1][j - 1]。

2、假设s是第k个长度为i的，包含不多于j个1的二进制串，那么，在长度为i + 1的，包含不多于j + 1个1的二进制串中，0s是第k个，1s是第(k + cnt[i][j])个，利用这个思路的逆向，我们就可以按顺序输出整个串。

/* ID: geterns1 PROG: kimbits LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; double cnt[33][33]; //cnt[i][j] save the number of the strings which include at most j 1s with length i void prtStr(int bits, int ones, double idx){ if (bits == 0){ printf("/n"); return; } if (idx > cnt[bits - 1][ones]){ printf("1"); prtStr(bits - 1, ones - 1, idx - cnt[bits - 1][ones]); } else{ printf("0"); prtStr(bits - 1, ones, idx); } } int main(){ freopen("kimbits.in", "r", stdin); freopen("kimbits.out", "w", stdout); int n, l; double idx; scanf("%d %d %lf", &n, &l, &idx); for (int i = 0; i <= n; i ++) for (int j = 0; j <= n; j ++) if (j == 0) cnt[i][j] = 1; else if (j > i) cnt[i][j] = cnt[i][i]; else cnt[i][j] = cnt[i - 1][j] + cnt[i - 1][j - 1]; prtStr(n, l, idx); fclose(stdin); fclose(stdout); return 0; }