2016

转载于 2017-08-13 20:39:00 发布 · 70 阅读

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原文链接：http://www.cnblogs.com/haoabcd2010/p/7354688.html

本文介绍了一道编程题的解决方案，题目要求计算满足特定条件的整数对数量，即两数乘积为2016的倍数。通过分析规律并运用编程技巧，提供了一个高效的算法实现。

1803: 2016

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

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

 1 # include <cstdio>
 2 # include <cstring>
 3 # include <cstdlib>
 4 # include <iostream>
 5 # include <vector>
 6 # include <queue>
 7 # include <stack>
 8 # include <map>
 9 # include <bitset>
10 # include <sstream>
11 # include <set>
12 # include <cmath>
13 # include <algorithm>
14 #pragma comment(linker,"/STACK:102400000,102400000")
15 using namespace std;
16 #define LL          long long
17 #define lowbit(x)   ((x)&(-x))
18 #define PI          acos(-1.0)
19 #define INF         0x3f3f3f3f
20 #define eps         1e-8
21 #define MOD         1000000007
22 
23 inline int scan() {
24     int x=0,f=1; char ch=getchar();
25     while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
26     while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
27     return x*f;
28 }
29 inline void Out(int a) {
30     if(a<0) {putchar('-'); a=-a;}
31     if(a>=10) Out(a/10);
32     putchar(a%10+'0');
33 }
34 #define MX 100005
35 /**************************/
36 LL a,b;
37 int main()
38 {
39     LL num=0;
40     for (int i=1;i<=2016;i++)
41         for (int j=1;j<=2016;j++)
42                 if (i*j%2016==0) num++;
43 
44     while (scanf("%lld%lld",&a,&b)!=EOF)
45     {
46         LL ca = a/2016;
47         LL ya = a%2016;
48         LL cb = b/2016;
49         LL yb = b%2016;
50         LL ans=0;
51 
52         ans += num*ca*cb;
53 
54         LL tp=0;
55         for (int i=1;i<=ya;i++)
56             for (int j=1;j<=2016;j++)
57             if (i*j%2016==0) tp++;
58         tp*=cb;
59 
60         for (int i=1;i<=ya;i++)
61             for (int j=1;j<=yb;j++)
62             if (i*j%2016==0) tp++;
63         ans = tp+ans;
64 
65         tp=0;
66         for (int i=1;i<=yb;i++)
67             for (int j=1;j<=2016;j++)
68             if (i*j%2016==0) tp++;
69         tp*=ca;
70 
71         ans = tp+ans;
72 
73         printf("%lld\n",ans);
74     }
75     return 0;
76 }