題意
Sol
啊啊這題好惡心啊，推的時候一堆細節qwq
(a \% i = a – frac{a}{i} * i)
把所有的都展開，直接分塊。關鍵是那個(i
ot= j)的地方需要減。。。。
然後就慢慢寫就好了
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, mod = 19940417, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline LL add(A x, B y) {
if(x + y < 0) return x + y + mod;
return x + y >= mod ? x + y - mod : x + y;
}
template <typename A, typename B> inline void add2(A &x, B y) {
if(x + y < 0) x = x + y + mod;
else x = (x + y >= mod ? x + y - mod : x + y);
}
template <typename A, typename B> inline LL mul(A x, B y) {
x = (x + mod) % mod;
y = (y + mod) % mod;
return 1ll * x * y % mod;
}
template <typename A> inline LL sqr(A x) {
return 1ll * x * x;
}
int N, M, a, b;
int sum(int l, int r) {
if(l == r) return l;
int n = r - l + 1;
if(n & 1) return add(mul(l, n), mul(n, (n - 1) / 2));
else return add(mul(l, n), mul(n / 2, n - 1));
}
int calc(int n) {
int ret = 0;
for(int i = 1, j; i <= n; i = j + 1) {
j = n / (n / i);
add2(ret, mul(n / j, sum(i, j)));
}
return ret;
}
int get(int x) {
int a = x, b = 2 * x + 1, c = x + 1;
if(a % 2 == 0) a /= 2;
else if(b % 2 == 0) b /= 2;
else if(c % 2 == 0) c /= 2;
if(a % 3 == 0) a /= 3;
else if(b % 3 == 0) b /= 3;
else if(c % 3 == 0) c /= 3;
return mul(mul(a, b), c);
}
int fuck2(int i, int j) {//sum k^2
return add(get(j), -get(i - 1));
}
int calc2() {
int ret = 0;
for(int i = 1, j; i <= N; i = j + 1) {
j = min(M / (M / i), N / (N / i));
int a = M / i, b = N / i;
add2(ret, add(add(mul(N, mul(a, sum(i, j))), mul(M, mul(b, sum(i, j)))), -mul(mul(a, b), fuck2(i, j))));
}
return ret;
}
signed main() {
cin >> N >> M;
if(N > M) swap(N, M);
a = calc(N);
b = calc(M);
int ans = mul(add(mul(N, N), -a), add(mul(M, M), -b));
add2(ans, -mul(N, mul(N, M)));
add2(ans, calc2());
cout << ans;
return 0;
}