A. Alyona and Numbers

題目連接：

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

題意

給你n,m。然后告訴你1<=x<=n,1<=y<=m

然后問你(x+y)%5=0的方案有多少種

題解：

考慮余數(shù)。

兩個余數(shù)之和為0，那么有0+0,1+4,2+3,3+2,4+1這么五種組合，我可以O(shè)(n)或者O(5)統(tǒng)計(jì)出每個數(shù)的余數(shù)為i的有多少個。

然后再O(5)的求解答案就好了。

代碼

#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; }

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