A. Alyona and Numbers【水题】-优快云博客

博客围绕一个编程问题展开，给定两个整数 n 和 m，分别代表两列从 1 到 n 和 1 到 m 的整数。需要计算从两列中各选一个整数，使它们的和能被 5 整除的对数。输入为 n 和 m 的值，输出为满足条件的对数，可通过看余数组队凑 5 来解决。

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Problem - 682A - Codeforces

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

Copy

6 12

output

Copy

input

Copy

11 14

output

Copy

input

Copy

1 5

output

Copy

input

Copy

3 8

output

Copy

input

Copy

5 7

output

Copy

input

Copy

21 21

output

Copy

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.

直接看余数，组队凑成5

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<deque>
#include<cmath>
#include<string.h>
using namespace std;
// ctrl+shift+C 注释
//ctrl+shift+x 取消
#define int long long
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;
typedef pair<int,int> PII;
const int N=2e5+10;
const ll M=1e18+10;
const int mod=1e9+7;
int a[N],sum[N];
priority_queue<int,vector<int>,greater<int> >pq;
set<int>se;
map<int,int>mp,mpp;
queue<int>qu;
vector<int>v;
deque<int>de;
void solve()
{
    int n,m;
    int sum=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        mp[i%5]++;
    }
    for(int i=1;i<=m;i++)
    {
        mpp[i%5]++;
    }
    for(int i=0;i<=n;i++)
    {
        if(i==0)
        {
            sum+=mp[0]*mpp[0];
        }
        else
        sum+=mp[i]*mpp[5-i];
    }
    cout<<sum<<endl;
}
signed main()
{
   int t=1;
   //cin>>t;
   while(t--)
   {
       solve();
   }
}