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
", ans);
}
return 0;
}