codeforces 682 a 解题报告 Alyona and Numbers 数论基础

题目：

A. Alyona and Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

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.

思路：

暴力两重循环超时了......

自己没想到优化的方法......

后来参考了juju的方法才A掉.......

附链接：http://www.cnblogs.com/Sunshine-tcf/p/5690183.html

代码：

#include<stdio.h>#include<iostream>#include<math.h>#include<algorithm>#include<limits.h>#include<string.h>using namespace std;int main(){int n,m,a[6],b[6];__int64 sum;while(~scanf("%d%d",&n,&m)){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];sum=(__int64)a[0]*(__int64)b[0];//int 和 __int64 之间需要显示类型转换 也可以直接把a b 数组开成 __int64 型 for(int i=1;i<5;++i){sum+=(__int64)a[i]*(__int64)b[5-i];}printf("%I64d\n",sum);}return 0;}