Codeforces Round #358 (Div. 2) A. Alyona and Numbers

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

题意：1<=a<=n,1<=b<=m,求a+b能整除5的方案数

思路：直接枚举5的倍数就好了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int a=min(n,m),b=max(n,m);
ll ans=0;
for(int i=5;i<=n+m;i+=5)
{
if(i<=a)
ans+=i-1;
else if(i>a&&i<=b)
ans+=a;
else
ans+=(a-i+b+1);
}
printf("%I64d\n",ans);
}
return 0;
}