[BZOJ2154]Crash的数字表格 && 莫比乌斯反演

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orz PoPoQQQ 课件上的例题啊orzorz

话说这种根号划分的方法好像次次都有的样子orzorz

http://wenku.baidu.com/link?url=RRtdDApIUqzKmUDS4IOPU0MQnOLMJ6xXv0hFSajFA05YkoiGQdrN6koBLNMrmxnMs0MXXRL_ELxQfwSOvVdtSsNLaKao96PtBY63tJai903

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<queue>#define SF scanf#define PF printfusing namespace std;typedef long long LL;const int MAXN = 10000000;const int MOD = 20101009;int mu[MAXN+10], prime[MAXN+10], S[MAXN+10];int n, m, tot;bool vis[MAXN+10];void init() {    mu[1] = 1;    for(int i = 2; i <= n; i++) {        if(!vis[i]) prime[++tot] = i, mu[i] = -1;        for(int j = 1; 1LL * prime[j] * i <= n; j++) {            vis[prime[j]*i] = true;            if(i % prime[j] == 0) {                mu[prime[j]*i] = 0;                break;            }            mu[prime[j]*i] = -mu[i];        }    }    for(int i = 1; i <= n; i++)         S[i] = (S[i-1] + 1LL * i * i * mu[i] % MOD) % MOD;}LL sum(LL x, LL y) {    return (x * (x+1) / 2) % MOD * (y * (y+1) / 2 % MOD) % MOD;}LL calc(int x, int y) {    LL ret = 0, Next_pos;    if(x > y) swap(x, y);    for(int i = 1; i <= x; i = Next_pos+1) {        Next_pos = min(x / (x/i), y / (y/i));        ret = (ret + (S[Next_pos] - S[i-1]) * sum(x/i, y/i) % MOD) % MOD;    }    return ret;}int main() {    SF("%d%d", &n, &m);    if(n > m) swap(n, m);    init();    LL Next_pos, ans = 0;    for(int i = 1; i <= n; i = Next_pos+1) {        Next_pos = min(n / (n/i), m / (m/i));        ans = (ans + (i+Next_pos) * (Next_pos-i+1) / 2 % MOD * calc(n/i, m/i) % MOD) % MOD;    }    cout << (ans+MOD)%MOD;}

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