Codeforces Round #358 (Div. 2) A. Alyona and Numbers 水题-优快云博客

本题探讨了如何通过统计余数来快速计算两个整数范围内所有可能配对中，其和能被5整除的配对数量。通过将问题简化为余数的组合问题，并利用O(n)的时间复杂度来解决。

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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

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;
}