类欧几里得算法推导

第一种

$$f(a,b,c,n)=\sum _{i=0}^n\frac{ai+b}{c}$$
情况一：$a\ge corb\ge c$
$$f(a,b,c,n)=f(a\%c,b\%c,c,n)+\frac{n(n+1)}{2}\cdot \frac{a}{c}+\frac{b}{c}\cdot (n+1)$$
情况二：$a<banda<c$
$$f(a,b,c,n)=\sum _{i=0}^n\sum _{j=1}^m[ai+b\ge cj]=(n+1)m-\sum _{j=0}^{m-1}\frac{cj+(c-b+a-1)}{a}=(n+1)m-f(c,c-b+a-1,a,m-1)$$

第二种

$$g(a,b,c,n)=\sum _{i=0}^{n}i\frac{ai+b}{c}$$
情况一：$a\ge corb\ge c$
$$g(a,b,c,n)=g(a\%c,b\%c,c,n)+\frac{n(n+1)(2n+1)}{6}\cdot \frac{a}{c}+\frac{n(n+1)}{2}\cdot \frac{b}{c}$$
情况二：$a<banda<c$
$$g(a,b,c,n)=\sum _{i=0}^{n}i\sum _{j=1}^m[ai+b\ge cj]=$$
$$\frac{1}{2}\left(mn(n+1)-\sum _{j=0}^{m-1}{\left(\frac{cj+(c-b+a-1)}{a}\right)}^2+\sum _{j=0}^{m-1}\frac{cj+(c-b+a-1)}{a}\right)=$$
$$\frac{1}{2}(mn(n+1)-h(c,c-b+a-1,a,m-1)+f(c,c-b+a-1,a,m-1))$$

第三种

$$h(a,b,c,n)=\sum _{i=0}^n{\left(\frac{ai+b}{c}\right)}^2$$
情况一：$a\ge corb\ge c$
$$h(a,b,c,n)=\frac{n(n+1)(2n+1)}{6}\cdot {\left(\frac{a}{c}\right)}^2+{\left(\frac{b}{c}\right)}^2\cdot (n+1)+\frac{a}{c}\cdot \frac{b}{c}n(n+1)+$$
$$2\frac{a}{c}g(a\%c,b\%c,c,n)+2\frac{b}{c}f(a\%c,b\%c,c,n)+h(a\%c,b\%c,c,n)$$
情况二：$a<banda<c$
$$h(a,b,c,n)=\sum _{i=0}^n\sum _{j=1}^m[ai+b\ge cj](2j-1)=\sum _{j=0}^{m-1}(2j+1)\left(n+1-\frac{cj+(c-b+a-1)}{a}\right)=$$
$$m^2(n+1)-2g(c,c-b+a-1,a,m-1)-f(c,c-b+a-1,a,m-1)$$

大板子

洛谷P5170

#include 
using namespace std;
typedef long long ll;

ll modpow(ll a, int b);
const int mod = 998244353, inv2 = (mod + 1) / 2, inv6 = modpow(6, mod - 2);
ll modpow(ll a, int b) {
    ll res = 1;
    for (; b; b >>= 1) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
    }
    return res;
}
struct Result { ll f, g, h; };
Result extgcd(ll a, ll b, ll c, ll n) {
    if (a == 0) return (Result) { b / c * (n + 1) % mod,
        n * (n + 1) / 2 % mod * (b / c) % mod,
        (b / c) * (b / c) % mod * (n + 1) % mod };
    if (a >= c || b >= c) {
        Result lst = extgcd(a % c, b % c, c, n);
        ll ta = n * (n + 1) / 2 % mod, tb = n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
        ll ac = a / c, bc = b / c;
        return (Result) { (ta * ac + (n + 1) * bc + lst.f) % mod,
            (tb * ac + ta * bc + lst.g) % mod,
            (tb * ac % mod * ac + (n + 1) * bc % mod * bc + ta * 2 * ac % mod * bc + 2 * lst.g * (a / c) + 2 * lst.f * (b / c) + lst.h) % mod };
    }
    ll m = (a * n + b) / c;
    Result lst = extgcd(c, c - b + a - 1, a, m - 1);
    return (Result) { ((n + 1) * m - lst.f + mod) % mod,
        (m * n % mod * (n + 1) + lst.f - lst.h + 2ll * mod) % mod * inv2 % mod,
        (m * m % mod * (n + 1) - 2 * lst.g - lst.f + 3ll * mod) % mod };
}
int main() {
    int T, a, b, c, n;
    for (scanf("%d", &T); T--;) {
        scanf("%d%d%d%d", &n, &a, &b, &c);
        Result res = extgcd(a, b, c, n);
        printf("%lld %lld %lld\n", res.f, res.h, res.g);
    }
    return 0;
}