一类求和问题——类欧几里得

转载自：https://zhuanlan.zhihu.com/p/34650451

今天要来介绍的是用类欧几里得算法来解决一类求和问题。

模板题

给出 $n, a, b, c$，对于每组数据，分别输出 $f, h, g$ 的值，答案对 $998244353$ 取模。（$n leq 10^9$）

//由于这三个函数是互相依赖的，所以我们将其放在一个函数里求解

//分别算会产生大量的重复计算

#include <cstdio>
#include <iostream>

using namespace std;
const long long inv2 = 499122177;
const long long inv6 = 166374059;
const long long mod = 998244353;
int t;
long long n, a, b, c;
struct query
{
    long long f;
    long long g;
    long long h;
};

long long read()
{
    long long ans = 0; char ch;
    while(!isdigit(ch = getchar()));
    while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + (ch ^ 48), ch = getchar();
    return ans;
}
query solve(long long a, long long b, long long c, long long n)
{
    query ans, prec;
    if(a == 0)
    {
        ans.f = (b / c) * (n + 1) % mod;
        ans.g = (b / c) * n % mod * (n + 1) % mod * inv2 % mod;
        ans.h = (b / c) * (b / c) % mod * (n + 1) % mod;
    }
    else if(a >= c || b >= c)
    {
        prec = solve(a % c, b % c, c, n);
        ans.f = (prec.f + n * (n + 1) % mod * inv2 % mod * (a / c) % mod + (n + 1) * (b / c) % mod) % mod;
        ans.g = ((a / c) * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod + (b / c) * n % mod * (n + 1) % mod * inv2 % mod + prec.g) % mod;
        ans.h = (prec.h + (a / c) * (a / c) % mod * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod + (n + 1) * (b / c) % mod * (b / c) % mod + 2 * (a / c) % mod * prec.g % mod + 2 * (b / c) % mod * prec.f % mod + 2 * (a / c) % mod * (b / c) % mod * n % mod * (n + 1) % mod * inv2 % mod) % mod;
    }
    else
    {
        long long m = (a * n + b) / c;
        prec = solve(c, c - b - 1, a, m - 1);
        ans.f = (n * (m % mod) % mod - prec.f) % mod;
        ans.g = (n * (n + 1) % mod * (m % mod) % mod - prec.f - prec.h) % mod * inv2 % mod;
        ans.h = (n * (m % mod) % mod * ((m + 1) % mod) % mod - 2 * prec.g - 2 * prec.f - ans.f) % mod;
    }
    return ans;
}
int main()
{
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld %lld %lld %lld", &n, &a, &b, &c);
        query ans = solve(a, b, c, n);
        printf("%lld %lld %lld
", (ans.f + mod) % mod, (ans.h + mod) % mod, (ans.g + mod) % mod);
    }
    return 0;
}

求f

$displaystyle f(a, b, c, n) =   sum_{i=0}^nleft lfloor frac{ai+b}{c} ight floor$

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

typedef long long ll;
const ll mod = 1e9+7;

ll f(ll a, ll b, ll c, ll n)
{
    if(!a)  return (b/c)*(n+1)%mod;
    if(a >= c || b >= c)  return (f(a%c, b%c, c, n) + n*(n+1)/2%mod*(a/c)%mod + (n+1)*(b/c)%mod) % mod;
    ll m = (a*n+b) / c;
    return (n*m%mod - f(c, c-b-1, a, m-1) + mod)%mod;
}

int main()
{
    printf("%lld
", f(1, 0, 2, 5));  //0+1+1+2+2=6
}