codeforces235E

题意

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求

∑i=1a∑j=1b∑k=1cd(ijk)

其中，d(i) 表示 i 的约数个数。

a,b,c≤2000

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Solution

设

f(i)=∑j=1a∑k=1b[j×k==i]

则

====∑i=1a∑j=1b∑k=1cd(ijk)∑i=1abf(i)×∑j=1cd(ij)∑i=1abf(i)×∑j=1c∑u|i∑v|j[(u,v)==1]∑u=1ab∑v=1c[(u,v)==1]∑u|iab×f(i)×⌊cv⌋∑u=1ab∑d|uμ(d)×∑d|v⌊cv⌋×∑u|iab×f(i)

然后我们发现全部都可以预处理出来！！！

因为这些枚举倍数的部分都是独立的，而且单单枚举倍数是可以做到O(nlogn)的。

所以时间复杂度就为O(ablogab)了，要卡常！

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Code

#include <bits/stdc++.h>typedef long long LL;#define FOR(i, a, b) for (int i = (a), i##_END_ = (b); i <= i##_END_; i++)#define DNF(i, a, b) for (int i = (a), i##_END_ = (b); i >= i##_END_; i--)template <typename Tp> void in(Tp &x) {    char ch = getchar(); x = 0;    while (ch < '0' || ch > '9') ch = getchar();    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();}template <typename Tp> Tp chkmax(Tp &x, Tp y) {return x > y ? x : x=y;}template <typename Tp> Tp chkmin(Tp &x, Tp y) {return x < y ? x : x=y;}template <typename Tp> Tp Max(const Tp &x, const Tp &y) {return x > y ? x : y;}template <typename Tp> Tp Min(const Tp &x, const Tp &y) {return x < y ? x : y;}const int MAXN = 4000010, MOD = (1 << 30);int a, b, c;int prime[MAXN];bool is_prime[MAXN];int h[MAXN], f[MAXN], q[MAXN], g[MAXN], ans, miu[MAXN];void get_prime(){    miu[1] = 1;    FOR(i, 2, a * b) {        if (!is_prime[i]) {            prime[++prime[0]] = i;            miu[i] = -1;        }        for (int j = 1; prime[j] * i <= i_END_; j++) {            is_prime[prime[j] * i] = true;            if (i % prime[j] == 0) {                miu[prime[j] * i] = 0;                break;            }            miu[prime[j] * i] = -miu[i];        }    }}int main(){    in(a); in(b); in(c); get_prime();    FOR(i, 1, c) for (int j = i; j <= c; j += i) h[i] = (h[i] + c / j) % MOD;    FOR(i, 1, a) FOR(j, 1, b) f[i * j]++;    FOR(i, 1, a * b) for (int j = i; j <= i_END_; j += i)        q[i] = (q[i] + f[j]) % MOD;    FOR(i, 1, a * b) for (int j = i; j <= i_END_; j += i)        g[j] = (g[j] + 1ll * miu[i] * h[i] % MOD) % MOD;    FOR(i, 1, a * b) g[i] = 1ll * g[i] * q[i] % MOD;    FOR(i, 1, a * b) ans = (ans + g[i]) % MOD;    printf("%d\n", ans);    return 0;}