[BZOJ2154]Crash的數字表格 && 莫比烏斯反演

orz PoPoQQQ 課件上的例題啊orzorz

話說這種根號劃分的方法好像次次都有的樣子orzorz

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

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
using 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;
}