CodeForces 682AAlyona and Numbers（被整除的个数）

http://codeforces.com/problemset/problem/682/A

A. Alyona and Numbers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.

Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and  equals 0.

As usual, Alyona has some troubles and asks you to help.

Input

The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).

Output

Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.

Examples input 6 12 output 14 input 11 14 output 31 input 1 5 output 1 input 3 8 output 5 input 5 7 output 7 input 21 21 output 88 Note

Following pairs are suitable in the first sample case:

for x = 1 fits y equal to 4 or 9; for x = 2 fits y equal to 3 or 8; for x = 3 fits y equal to 2, 7 or 12; for x = 4 fits y equal to 1, 6 or 11; for x = 5 fits y equal to 5 or 10; for x = 6 fits y equal to 4 or 9.

Only the pair (1, 4) is suitable in the third sample case.

题意：

给定两组数字，随机各取出一个数字，问一共有多少取法。

思路：

对 n 或者 m 进行一次从 1 开始的遍历，例如对于 1~m 有

<strong>ans=ans+(n+j)/5-j/5;// </strong>注意数据类型的溢出问题 ！

Code：

#include<cstdio> #include<cstring> const int MYDD=1103; int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { double ans=0.0; for(int j=1; j<=m; j++) { ans=ans+(n+j)/5-j/5;// } printf("%.0lf\n",ans); } return 0; } /* By:Shyazhut */