BZOJ2956: 模积和——整除分块

题意

求 $sum_{i=1}^n sum_{j=1}^m (n mod i)*(m mod j)$（$i eq j$），$n,m leq 10^9$答案对 $19940417$ 取模。

分析：

由于取模可化成取整的形式，$k mod i = k - left lfloor frac{k}{i} ight floor * i$，详见 BZOJ1257 余数之和。

易知，$sum_{i=1}^n sum_{j=1}^m (n mod i)*(m mod j) = sum_{i=1}^n(n mod i)sum_{j=1}^m(m mod j)$

所以答案为两部分余数和的乘积减去 $i$ 等于 $j$ 的情况，

当 $i=j$ 时，
$$
egin{aligned}
sum_{i=1}^{min(n,m)}(n mod i)(m mod i)  & = sum_{i=1}^{min(n,m)}(n - left lfloor frac{n}{i}  ight  floor  i)(m - left lfloor frac{m}{i} ight floor  i) \
&= sum_{i=1}^{min(n,m)}(nm - mleft lfloor frac{n}{i} ight floor  i - nleft lfloor frac{m}{i} ight floor i + left lfloor frac{n}{i} ight floor left lfloor frac{m}{i} ight floor i^2)  \
end{aligned}$$

代码：

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod = 19940417;
const ll inv6 = 3323403;
ll n, m;

//sum_1^n [k/i]*i
ll S1(ll n, ll k)
{
    ll ret = 0;
    if(k <= n)   //需要分类讨论
    {
        for(ll i = 1,j;i <= k;i = j+1)
        {
            j = k / (k / i);

            ret = (ret + (i+j) * (j-i+1) / 2 % mod * (k / i) % mod) % mod;
        }
    }
    else
    {
        for(ll i = 1,j;i <= n;i = j+1)
        {
            j = min(k / (k / i), n);
            ret =  (ret + (i+j) * (j-i+1) / 2 % mod * (k / i) % mod) % mod;
        }
    }

    return ret;
}

ll S2(ll n)
{
    n %= mod;
    return n * (n+1) % mod * (2*n+1) % mod * inv6 % mod;
}

//[n/i][m/i]i^2
ll S3(ll n, ll m)
{
    ll ret = 0;
    for(ll i = 1,j;i <= min(n, m);i = j+1)
    {
        j = min(n/(n/i), m/(m/i));
        ret = (ret + (n/i) * (m/i) % mod * (S2(j) - S2(i-1)) % mod) % mod;
    }
    return ret;
}

int main()
{
    scanf("%d%d", &n, &m);
    if(m > n) swap(n, m);
    ll ans = 1;
    ans = ans * (n*n%mod - S1(n, n) + mod) % mod;
    ans = ans * (m*m%mod - S1(m, m) + mod) % mod;
    ll ans2 = n *m % mod * m % mod;
    ans2 = (ans2 - m * S1(m, n) % mod) % mod;
    ans2 = (ans2 - n * S1(m, m) % mod) % mod;
    ans2 = (ans2 + S3(n, m)) % mod;
    //printf("%lld %lld
", ans, ans2);
    printf("%lld
", (ans - ans2 + 2*mod) % mod);
    return 0;
}

参考链接：https://www.cnblogs.com/henry-1202/p/10201032.html