@(POJ)[Stirling數, 排列組合, 數形結合]

Description

The Stirling number of the second kind S(n, m) stands for the number of ways to partition a set of n things into m nonempty subsets. For example, there are seven ways to split a four-element set into two parts:
{1, 2, 3} U {4}, {1, 2, 4} U {3}, {1, 3, 4} U {2}, {2, 3, 4} U {1}
{1, 2} U {3, 4}, {1, 3} U {2, 4}, {1, 4} U {2, 3}.
There is a recurrence which allows to compute S(n, m) for all m and n.
S(0, 0) = 1; S(n, 0) = 0 for n > 0; S(0, m) = 0 for m > 0;
S(n, m) = m S(n - 1, m) + S(n - 1, m - 1), for n, m > 0.
Your task is much "easier". Given integers n and m satisfying 1 <= m <= n, compute the parity of S(n, m), i.e. S(n, m) mod 2.
Example:
S(4, 2) mod 2 = 1.
Task
Write a program which for each data set:
reads two positive integers n and m,
computes S(n, m) mod 2,
writes the result.

Input

The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 200. The data sets follow.
Line i + 1 contains the i-th data set - exactly two integers ni and mi separated by a single space, 1 <= mi <= ni <= 10^9.

Output

The output should consist of exactly d lines, one line for each data set. Line i, 1 <= i <= d, should contain 0 or 1, the value of S(ni, mi) mod 2.

Sample Input

1 4 2

Sample Output

1

Solution

題意:
求斯特林數\[ \left\{ \begin{array}{} n \\ k \end{array}{} \right\} \% 2\]\[n, m \in [1, 10^9]\]
這題直接求解肯定是會T的, 因此考慮優化.

轉載自sdchr博客
侵刪

代碼附上:

#include<cstdio> #include<cctype> using namespace std;inline int read() {int x = 0, flag = 1;char c;while(! isdigit(c = getchar()))if(c == '-')flag *= - 1;while(isdigit(c))x = x * 10 + c - '0', c = getchar();return x * flag; }void println(int x) {if(x < 0)putchar('-'), x *= - 1;if(x == 0)putchar('0');int ans[1 << 5], top = 0;while(x)ans[top ++] = x % 10, x /= 10;for(; top; top --)putchar(ans[top - 1] + '0');putchar('\n'); }long long getQuantity(int x) {long long ret = 0;for(int i = 2; i <= x; i <<= 1)ret += x / i;return ret; }int calculate(int x, int y) {return getQuantity(x) - getQuantity(y) - getQuantity(x - y) == 0; }int main() {int T = read();while(T --){int n = read(), m = read();int d = n - m, oddQua = (m + 1) / 2;println(calculate(d + oddQua - 1, oddQua - 1));} }

轉載于:https://www.cnblogs.com/ZeonfaiHo/p/6444001.html

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