BZOJ2956: 模积和(数论分块)

题意

题目链接

Sol

啊啊这题好恶心啊，推的时候一堆细节qwq

(a \% i = a - frac{a}{i} * i)

把所有的都展开，直接分块。关键是那个(i ot= j)的地方需要减。。。。

然后就慢慢写就好了

#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;
}