Codeforces Round #319 (Div. 2) A. Multiplication Table | 书脊

Codeforces Round #319 (Div. 2) A. Multiplication Table | 书脊
http://codeforces.com/contest/577/problem/A

Let's consider a table consisting of n rows and n columns. The cell located at the intersection of i-th row and j-th column contains number i × j. The rows and columns are numbered starting from 1.

You are given a positive integer x. Your task is to count the number of cells in a table that contain number x.

题目大意: 给两个数字,n和k, 找出等于k的个数在n*n的table中, table中每一位是i*j, where i and j are row and column

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分析: 这个暴不了, 所以考的是观察. 观察题中给的example:

我们可以发现, 在一个n=6的表中, k=12 出现在6 4 3 2. 这些数字都是12的除数, 可是少了几个啊, 12本身也是除数,1 也是除数, 而且在n=10,k=5中, 我们发现答案是2(1和5). 那么上边的1和12 为什么不算? 因为 12/1 = 12 , 12不在表中. 所以我们得到另一个条件, 除完的余数要小于等于n.

public void solve(int testNumber, InputReader in, OutputWriter out) {

            int n = in.readInt();

            int k = in.readInt();

            int count = 0;

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

                if (k % i == 0 && k / i <= n)

                    count++;

            }

            out.print(count);

        }

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