BZOJ2956 模积和

2019-04-13 14:25发布

站内文章 / 模拟电子
15064 0 1086
[Solution] The original equation equals to: sigma(n % i) * sigma(m % j) - sigma((n % i) * (m % i)). That's the solution to it. You need to divide them into sqrt(n) parts by merging each equal (n % i) and (m % i). Another useful equation is that 1^2 + 2^2 + ... + n^2 = n * (n + 1) * (2 * n + 1) / 6. Be careful that 19940417 is not a prime number.
[Code] #include #include #include using namespace std; typedef long long qw; #ifdef WIN32 #define lld "%I64d" #else #define lld "%lld" #endif #define _l (qw) const int mod = 19940417; const int inv2 = 9970209; const int inv6 = 3323403; inline qw sqs(qw n) { return n * (n + 1) % mod * (n * 2 + 1) % mod * inv6 % mod; } inline qw sqn(qw n) { return (n + 1) * n % mod * inv2 % mod; } qw sovo(qw n) { qw s = n * n; for (qw i = 1, j; i <= n; i = j + 1) { j = n / (n / i); s -= (j - i + 1) * (i + j) / 2 * (n / i); } return s % mod; } qw sovt(qw n, qw m) { qw e = min(n, m); qw s = m * n % mod * e % mod; for (qw i = 1, j; i <= e; i = j + 1) { j = min(n / (n / i), m / (m / i)); s -= (n * (m / i) + m * (n / i)) % mod * (sqn(j) - sqn(i - 1)) % mod; s += (n / i) * (m / i) % mod * (sqs(j) - sqs(i - 1) + mod) % mod; s %= mod; if (s < 0) s += mod; } return s; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); #endif qw n, m, ans; scanf(lld lld, &n, &m); ans = (sovo(n) * sovo(m) % mod - sovt(n, m) + mod) % mod; printf("%d ", (int)ans); }

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