[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=2956

[算法]

首先有兩個重要的等式 :

1. 1 + 2 + 3 + 4 + ... + n = n(n + 1) / 2

2. 1 ^ 2 + 2 ^ 2 + ... + n ^ 2 = n(n + 1)(2n + 1) / 6

根據這兩個式子 ， 配合數論分塊 ， 即可在O(SQRT(N + M))的時間內求解此問題

[代碼]

#include<bits/stdc++.h>
using
 
 namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int P = 19940417;
const int PHI = 17091780;

int n , m , A , nxt , inv , inv2;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void
 
 chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline 
 
void sub(int &x , int y)
{
        x -= y;
        while (x < 0) x += P;
}
inline void add(int &x , int y)
{
        x += y;
        while (x >= P) x -= P;
}
inline int exp_mod(int a , int n)
{
        int b = a , res = 1;
        while (n > 0)
        {
                if (b & 1) res = 1LL * res * b % P;
                b = 1LL * b * b % P;
                n >>= 1;        
        }        
        return res;
}
inline int calc(int l , int r)
{
        int S1 = 1LL * r * (r + 1) % P * (2 * r + 1) % P * inv2 % P , 
                S2 = 1LL * (l - 1) * l % P * (2 * l - 1) % P * inv2 % P;
        return (S1 - S2 + P) % P;        
}

int main()
{
        
        read(n); read(m);
        inv = (P + 1) >> 1 , inv2 = 3323403;
        A = 1LL * m * m % P , nxt = 0;
        for (int i = 1; i <= m; i = nxt + 1)
        {
                nxt = m / (m / i);
                sub(A , 1LL * (m / i) * (i + nxt) % P * (nxt - i + 1) % P * inv % P);        
        }
        nxt = 0;
        int res = 1LL * n * n % P * A % P;
        for (int i = 1; i <= n; i = nxt + 1)
        {
                nxt = n / (n / i);
                sub(res , 1LL * A * (n / i) % P * (i + nxt) % P * (nxt - i + 1) % P * inv % P); 
        }
        if (n > m) swap(n , m);
        nxt = 0;
        int con = 1LL * n * n % P * m % P;
        for (int i = 1; i <= n; i = nxt + 1)
        {
                nxt = min(n / (n / i) , m / (m / i));
                add(con , 1LL * (n / i) * (m / i) % P * calc(i , nxt) % P); 
        }
        nxt = 0;
        for (int i = 1; i <= n; i = nxt + 1)
        {
                nxt = n / (n / i);
                sub(con , 1LL * (n / i) * m % P * (i + nxt) % P * (nxt - i + 1) % P * inv % P);
        }
        nxt = 0;
        for (int i = 1; i <= n; i = nxt + 1)
        {
              nxt = min(m / (m / i) , n);
                sub(con , 1LL * (m / i) * n % P * (i + nxt) % P * (nxt - i + 1) % P * inv % P);
        }
        sub(res , con);
        printf("%d\n" , res);
        
        return 0;
    
}

[BZOJ 2956] 模積和