C. Floor and Mod
令(⌊ab⌋=a \% b=k)，易推导得：若(x)满足条件，则(x=kb+k)。直接去枚举(k)，去构造(b)能取值的区间(lb)和(rb)，(ans+=rb-lb+1)。
先构造左区间的(lb=k+1),假设左区间合法；去构造右区间的(rb=b/k-1)，然后check左右区间是否满足(lb<=rb)，如果满足则计入答案，否则直接(break)。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define vll vector<ll>
#define vpll vector<pll>
#define fastio ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
double pi = acos(-1);
const double eps = 1e-9;
const int inf = 1e9 + 7;
const ll lnf = 1e18 + 7;
const int maxn = 1e5 + 10;
ll mod = 1e9 + 7;
int main()
{
fastio;
int t;
cin >> t;
while (t--)
{
ll a, b;
cin >> a >> b;
b = min(a, b);
ll ans = 0;
for (ll k = 1; k < b; k++)//目的是构造b
{
ll lb = k + 1, rb = a / k - 1;
//左区间的b，右区间的b
if (k > rb)break;
rb = min(rb, b);
ans += rb - lb + 1;
}
cout << ans << endl;
}
return 0;
}
D. Multiples and Power Differences
(720720 = 2 * 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 * 13 = lcm(1,2,...,16)) ，这样所有数的四次方与(720720)做差都是1 到 16的倍数。
让((i+j)\%2=0)的块为(720720)，其余为(720720-x^4)
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define vll vector<ll>
#define fastio ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
double pi = acos(-1);
const double eps = 1e-9;
const int inf = 1e9 + 7;
const ll lnf = 1e18 + 7;
const int maxn = 2e5 + 10;
ll mod = 1e9 + 7;
int main()
{
fastio;
//cout << 2 * 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 * 13 << endl;
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
int x;
cin >> x;
if ((i + j) & 1)
cout << 720720 - x*x*x*x << " ";
else cout << 720720<<" ";
}
cout << endl;
}
return 0;
}