[BZOJ3994][SDOI2015]约数个数和

3994: [SDOI2015]约数个数和

Time Limit: 20 Sec  Memory Limit: 128 MB Submit: 1104  Solved: 762 [Submit][Status][Discuss]

Description

 设d(x)为x的约数个数，给定N、M，求  

 

 

Input

输入文件包含多组测试数据。

第一行，一个整数T，表示测试数据的组数。

接下来的T行，每行两个整数N、M。

 

Output

 T行，每行一个整数，表示你所求的答案。

 

Sample Input

2
7 4
5 6

Sample Output

110
121

HINT

 1<=N, M<=50000

1<=T<=50000

 

图片好像挂了。。。那张图是$sum_{i=1}^nsum_{j=1}^m dleft(ij ight)$

如果没挂请忽视上面那排话

 

由对称性，不妨设$nle m$

有一个结论$dleft(xy ight)=sum_{imid x}sum_{jmid y}left[gcdleft(i,j ight)=1 ight]$

这个证明的话可以考虑每个质因数的贡献。。。意会一下

那么可以得到

$ans=sum_{x=1}^nsum_{y=1}^msum_{imid x}sum_{jmid y}left[gcdleft(i,j ight)=1 ight]$

$=sum_{i=1}^nsum_{j=1}^mlfloorfrac{n}{i} floorlfloorfrac{m}{j} floorleft[gcdleft(i,j ight)=1 ight]$

$=sum_{i=1}^nsum_{j=1}^mlfloorfrac{n}{i} floorlfloorfrac{m}{j} floorsum_{dmid i,dmid j}muleft(d ight)$

$=sum_{d=1}^nmuleft(d ight)sum_{i=1}^{lfloorfrac{n}{d} floor}sum_{j=1}^{lfloorfrac{m}{d} floor}lfloorfrac{n}{id} floorlfloorfrac{m}{id} floor$

$=sum_{d=1}^nmuleft(d ight)sum_{i=1}^{lfloorfrac{n}{d} floor}sum_{j=1}^{lfloorfrac{m}{d} floor}lfloorfrac{lfloorfrac{n}{d} floor}{i} floorlfloorfrac{lfloorfrac{m}{d} floor}{i} floor$

$=sum_{d=1}^nmuleft(d ight)left(sum_{i=1}^{lfloorfrac{n}{d} floor}lfloorfrac{lfloorfrac{n}{d} floor}{i} floor ight)left(sum_{j=1}^{lfloorfrac{m}{d} floor}lfloorfrac{lfloorfrac{m}{d} floor}{i} floor ight)$

令$gleft(n ight)=sum_{i=1}^nlfloorfrac{n}{i} floor$

那么$ans=sum_{d=1}^nmuleft(d ight)gleft(lfloorfrac{n}{d} floor ight)gleft(lfloorfrac{m}{d} floor ight)$

而$g$可以通过枚举每个分子然后不停的往倍数上加$1$，然后扫一遍前缀和求出，我为了用Latex码数学公式现在已经头昏眼花神志不清，如果你觉得我已经开始胡言乱语了导致你没看懂那就看看代码吧

预处理时间复杂度为$Oleft(nlnn ight)$

似乎神犇们都是$Oleft(n ight)$预处理？？？我还是太菜了哎

每次询问的话分块求，总时间复杂度$Oleft(Tsqrt{n} ight)$

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 50000 + 10;
bool mark[maxn] = {false};
int mu[maxn], g[maxn] = {0}, sum[maxn];
int pri[maxn], prn = 0;
void shai(){
    mu[1] = 1;
    for(int i = 2; i < maxn; i++){
        if(!mark[i]){
            mu[i] = -1;
            pri[++prn] = i;
        }
        for(int j = 1; j <= prn && pri[j] * i < maxn; j++){
            mark[i * pri[j]] = true;
            if(i % pri[j] == 0){
                mu[i * pri[j]] = 0;
                break;
            }
            else mu[i * pri[j]] = -mu[i];
        }
    }
    for(int i = 1; i < maxn; i++)
        for(int j = i; j < maxn; j += i)
            g[j]++;
    sum[0] = g[0] = 0;
    for(int i = 1; i < maxn; i++){
        sum[i] = mu[i] + sum[i - 1];
        g[i] += g[i - 1];
    }
}
int main(){
    shai();
    int T, n, m;
    ll ans;
    scanf("%d", &T);
    while(T--){
        scanf("%d %d", &n, &m);
        if(n > m) swap(n, m);
        ans = 0;
        for(int p, i = 1; i <= n; i = p + 1){
            p = min(n / (n / i), m / (m / i));
            ans += (ll) (sum[p] - sum[i - 1]) * g[n / p] * g[m / p];
        }
        printf("%lld
", ans);
    }
    return 0;
}