CodeForces 682A Alyona and Numbers （水题）

Alyona and Numbers

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#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

Input

6 12

Output

14

Input

11 14

Output

31

Input

1 5

Output

1

Input

3 8

Output

5

Input

5 7

Output

7

Input

21 21

Output

88

题意：

找出有多少数对(x, y), 1 ≤ x ≤ n, 1 ≤ y ≤ m 使得x+y能整除5；

题解：

直接模拟找出所有数模5的同余数；

注意long long；

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 110
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,m;
LL a[6],b[6];

int main(int argc, char const *argv[])
{
    //IN;

    while(scanf("%d %d",&n,&m) != EOF) {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));

        for(int i=1; i<=n; i++){
            a[i%5]++;
        }
        for(int i=1; i<=m; i++){
            b[i%5]++;
        }

        LL ans = a[0] * b[0];
        for(int i=1; i<5; i++) {
            ans += a[i]*b[5-i];
        }

        printf("%I64d\n", ans);
    }

    return 0;
}