BZOJ 2693: jzptab( 莫比乌斯反演 )

速度居然#2...目测是因为我没用long long..

求∑ lcm(i, j) (1 <= i <= n, 1 <= j <= m)

化简之后就只须求f(x) = x∑u(d)*d (d | x) 然后就是分块了...

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#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 10000009;const int MOD = 100000009; bool check[maxn];int f[maxn], prime[maxn], N = 0; void init() { memset(check, false, sizeof check); f[0] = 0; f[1] = 1; for(int i = 2; i < maxn; i++) { if(!check[i]) { prime[N++] = i; f[i] = ll(i) * (1 - i) % MOD; } for(int j = 0; j < N && ll(i) * prime[j] < maxn; j++) { check[i * prime[j]] = true; if(i % prime[j]) f[i * prime[j]] = ll(f[i]) * f[prime[j]] % MOD; else { f[i * prime[j]] = ll(prime[j]) * f[i] % MOD; break; } } } for(int i = 1; i < maxn; i++) f[i] = (f[i] + f[i - 1]) % MOD;} inline int sum(int a, int b) { return (ll(a) * (a + 1) / 2 % MOD) * (ll(b) * (b + 1) / 2 % MOD) % MOD;} void work(int x, int y) { if(x > y) swap(x, y); int ans = 0; for(int L = 1; L <= x; L++) { int R = min(x / (x / L), y / (y / L)); (ans += 1ll * sum(x / L, y / L) * (f[R] - f[L - 1]) % MOD) %= MOD; L = R; } printf("%d\n", (ans + MOD) % MOD);} int main() { init(); int t; scanf("%d", &t); while(t--) { int x, y; scanf("%d%d", &x, &y); work(x, y); } return 0;}-------------------------------------------------------------------

2693: jzptab

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 602 Solved: 237
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Description

Input

一个正整数T表示数据组数

接下来T行 每行两个正整数 表示N、M

Output

T行 每行一个整数 表示第i组数据的结果

Sample Input

1

4 5

Sample Output

122

HINT
T <= 10000

N, M<=10000000

HINT

Source

版权所有者： 倪泽堃