A. Alyona and Numbers
题目连接：
http://www.codeforces.com/contest/682/problem/A
Description
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.
Sample Input
6 12
Sample Output
14
Hint
题意
给你n,m。然后告诉你1<=x<=n,1<=y<=m
然后问你(x+y)%5=0的方案有多少种
题解：
考虑余数。
两个余数之和为0，那么有0+0,1+4,2+3,3+2,4+1这么五种组合，我可以O(n)或者O(5)统计出每个数的余数为i的有多少个。
然后再O(5)的求解答案就好了。
代码
#include<bits/stdc++.h>
using namespace std;
long long num1[5];
long long num2[5];
int main()
{
long long n,m;
cin>>n>>m;
for(int i=0;i<5;i++)
{
num1[i]=n/5;
if(n%5>=i)num1[i]++;
num2[i]=m/5;
if(m%5>=i)num2[i]++;
}
num1[0]--,num2[0]--;
long long ans = 0;
ans = num1[0]*num2[0]+num1[1]*num2[4]+num1[2]*num2[3]+num1[3]*num2[2]+num1[4]*num2[1];
cout<<ans<<endl;
}