Problem A: 2016

说实话，对于我这萌新来说真的难。。。。后面看了大牛的博客才懂的，就是n和m分别对2016取模，然后放到数组里存。

#include<iostream>#include<string.h>#include<stdlib.h>#include<math.h>#include <algorithm>#pragma warning(disable:4996)using namespace std;long long a[2030], b[2030];int main(){long long n, m;int i, j;long long k, t;long long count;while (~scanf("%lld%lld", &n, &m)){count = 0;k = n / 2016;t = n % 2016;for (i = 1; i <= t; i++)a[i] = k + 1;for (i = t + 1; i <= 2016; i++)a[i] = k;k = m / 2016;t = m % 2016;for (i = 1; i <= t; i++)b[i] = k + 1;for (i = t + 1; i <= 2016; i++)b[i] = k;for (i = 1; i <= 2016; i++){for (j = 1; j <= 2016; j++)if ((i*j) % 2016 == 0)count = cou + a[i] * b[j];}printf("%lld\n",count);}return 0;}

以下的大牛代码，然而看不懂。。。。

#include<iostream>#include<string.h>#include<stdlib.h>#include<math.h>#include <algorithm>#pragma warning(disable:4996)using namespace std;int n, m;long long f(int g){int nn = n / g;long long s = nn - (1008 % g == 0)*nn / 2 - (672 % g == 0)*nn / 3 - (288 % g == 0)*nn / 7;s += (336 % g == 0)*nn / 6 + (96 % g == 0)*nn / 21 + (144 % g == 0)*nn / 14 - (48 % g == 0)*nn / 42;long long gg = g;return gg*m / 2016 * s;}long long f7(int g){return f(g) + f(g * 7);}long long f5(int g){return f7(g) + f7(g * 3) + f7(g * 9);}int main(){while (cin >> n >> m)cout << f5(1) + f5(2) + f5(4) + f5(8) + f5(16) + f5(32) << endl;return 0;}