莫比乌斯繁衍1

bzoj crash的数字表格 2154

题目大意

n , m <= 10^7

终于会正经的莫比乌斯繁衍了。就是包含两种变换1.莫比乌斯繁衍、莫比乌斯变换 2.考虑贡献（被计算的次数），进行枚举顺序的变化，从而达到线性筛的目的。

因为过程太繁琐，用图片替代输入

减少取模次数有奇效，不知道为什么如此慢

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 10000020
#define mod 20101009

int prime[maxn],cnt;
long long f[maxn];
bool tag[maxn];
long long n,m,ans;

void init(){
    f[1] = 1;
    for (register int i = 2 ; i <= n ; i++){
        if ( !tag[i] ) prime[++cnt] = i , f[i] = 1 - i;
        for (register int j = 1 ; j <= cnt && prime[j] * i <= n ; j++){
            tag[i * prime[j]] = 1;
            if ( (i % prime[j]) == 0 ){
                f[i * prime[j]] = f[i];
                break;
            }
            else{
                f[i * prime[j]] = (f[i] * (long long) (1 - prime[j]));
            }
        }
    }
    for (register long long i = 1 ; i <= n ; i++) f[i] = (f[i] * i);
    for (register int i = 2 ; i <= n ; i++) f[i] = (f[i] + f[i - 1]) % mod;
}
inline long long power(long long x,long long y){
    long long res = 1;
    while ( y ) {
        if ( y & 1 ) res = (res * x) % mod;
        x = (x * x) % mod;
        y >>= 1;
    }
    return res % mod;
}
int main(){
    freopen("input.txt","r",stdin);
    scanf("%lld %lld",&n,&m);
    if ( n > m ) swap(n,m);
    init();
    for (register long long i = 1 ; i <= min(n,m) ;){
        long long next = min(n/(n/i),m/(m/i));
        ans = (ans + (((f[next] - f[i - 1]) + mod) % mod * (n/i) % mod * (m/i) % mod * (n/i + 1) % mod * (m/i + 1) % mod)) % mod;
        i = next + 1;
    }
    ans = (ans * power(4,mod - 2)) % mod;
    printf("%d
",(int)((ans % mod + mod) % mod));
}