湖南省第十二届大学生计算机程序设计竞赛 problem A 2016

如果 a * b % 2016 == 0

如果a = 1 ,且 a * b % 2016 == 0

考虑一下a = 2017的时候

2017 * b = (2016 + 1) * b % 2016 == 0必定成立

那么就是说1中搭配成的b，2017一样能搭配。

同样:4033 * b = (2016 + 2016 + 1) * b % 2016 == 0必定成立

所以，我可以枚举[1,2016]中[1,2016]中，i * j % 2016 == 0的对数，然后乘上对应的[1,n]中有i这个数的个数，代替数也算，代替数就是那些等价数，1 --- 2017 --- 4033 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
LL n, m;
const int maxn = 2016 + 20;
LL numn[maxn], numm[maxn];
void work () {
   LL tnk = n / 2016;
   LL rn =  n % 2016;
   memset (numn, 0, sizeof numn);
   memset (numm, 0, sizeof numm);
   for (int i = 1; i <= rn; ++i) {
       numn[i] = tnk + 1;
   }
   for (int i = rn + 1; i <= 2016; ++i) {
       numn[i] = tnk;
   }
   LL tmk = m / 2016;
   LL rm = m % 2016;
   for (int i = 1; i <= rm; ++i) {
       numm[i] = tmk + 1;
   }
   for (int i = rm + 1; i <= 2016; ++i) {
       numm[i] = tmk;
   }
   LL ans = 0;
   for (int i = 1; i <= 2016; ++i) {
       for (int j = 1; j <= 2016; ++j) {
           if ((i * j) % 2016 == 0) {
               ans += numn[i] * numm[j];
           }
       }
   }
   cout << ans << endl;
}
int main () {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    while (cin >> n >> m) {
        work ();
    }
    return 0;
}