Codeforces Round #319 (Div. 2) B. Modulo Sum | 书脊
http://codeforces.com/contest/577/problem/B
题目大意: 给n个数, 问能不能找到其中的subsequence的合可以整除k.
看了答案后, 感觉思路差不多, 首先,当n>k时, 因为鸽巢原理, n个数,每个贡献1,肯定能找到合整除k的. 而当n<=k时, 我们可以通过记录数组当前数字与前一个数组合的模来判断是不是是否可以整除, 即: 如果合的模是0, 那么找到了答案返回YES.
Read full article from Codeforces Round #319 (Div. 2) B. Modulo Sum | 书脊
http://codeforces.com/contest/577/problem/B
You are given a sequence of numbers a1, a2, ..., an, and a number m.
Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible bym.
input
3 5 1 2 3
output
YES
input
1 6 5
output
NO
input
4 6 3 1 1 3
output
YES
input
6 6 5 5 5 5 5 5
output
YES
Note
In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
In the third sample test you need to choose two numbers 3 on the ends.
In the fourth sample test you can take the whole subsequence.
看了答案后, 感觉思路差不多, 首先,当n>k时, 因为鸽巢原理, n个数,每个贡献1,肯定能找到合整除k的. 而当n<=k时, 我们可以通过记录数组当前数字与前一个数组合的模来判断是不是是否可以整除, 即: 如果合的模是0, 那么找到了答案返回YES.
Read full article from Codeforces Round #319 (Div. 2) B. Modulo Sum | 书脊
