2016

1803: 2016

Time Limit: 5 Sec     Memory Limit: 128 Mb     Submitted: 1506     Solved: 855    

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Description

 给出正整数 n 和 m，统计满足以下条件的正整数对 (a,b) 的数量：

 

1. 1≤a≤n,1≤b≤m;

2. a×b 是 2016 的倍数。

 

Input

输入包含不超过 30 组数据。

每组数据包含两个整数 n,m (1≤n,m≤109).

 

Output

对于每组数据，输出一个整数表示满足条件的数量。

 

Sample Input

32 63
2016 2016
1000000000 1000000000

 

Sample Output

1
30576
7523146895502644

 

Hint

 

Source

湖南省第十二届大学生计算机程序设计竞赛

//叉姐还是厉害啊，一道去年的省赛水题我想半年，还是太弱了！

题解: 如果 a*b = 2016 那么，((a+2016) * b)%2016 = 0 (a * (b+2016)) %2016 = 0 所以找着规律再套一下数据，基本写出来就是必过的

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <sstream>
# include <set>
# include <cmath>
# include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")
using namespace std;
#define LL          long long
#define lowbit(x)   ((x)&(-x))
#define PI          acos(-1.0)
#define INF         0x3f3f3f3f
#define eps         1e-8
#define MOD         1000000007

inline int scan() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
#define MX 100005
/**************************/
LL a,b;
int main()
{
    LL num=0;
    for (int i=1;i<=2016;i++)
        for (int j=1;j<=2016;j++)
                if (i*j%2016==0) num++;

    while (scanf("%lld%lld",&a,&b)!=EOF)
    {
        LL ca = a/2016;
        LL ya = a%2016;
        LL cb = b/2016;
        LL yb = b%2016;
        LL ans=0;

        ans += num*ca*cb;

        LL tp=0;
        for (int i=1;i<=ya;i++)
            for (int j=1;j<=2016;j++)
            if (i*j%2016==0) tp++;
        tp*=cb;

        for (int i=1;i<=ya;i++)
            for (int j=1;j<=yb;j++)
            if (i*j%2016==0) tp++;
        ans = tp+ans;

        tp=0;
        for (int i=1;i<=yb;i++)
            for (int j=1;j<=2016;j++)
            if (i*j%2016==0) tp++;
        tp*=ca;

        ans = tp+ans;

        printf("%lld
",ans);
    }
    return 0;
}