第一种
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f(a,b,c,n)=\sum_{i=0}^n\frac{ai+b}{c}
f(a,b,c,n)=i=0∑ncai+b
情况一：
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a\ge c~or~b \ge c
a≥c or b≥c
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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)
f(a,b,c,n)=f(a%c,b%c,c,n)+2n(n+1)⋅ca+cb⋅(n+1)
情况二：
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a
n
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a<b~and~a<c
a<b and a<c
f
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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)
f(a,b,c,n)=i=0∑nj=1∑m[ai+b≥cj]=(n+1)m−j=0∑m−1acj+(c−b+a−1)=(n+1)m−f(c,c−b+a−1,a,m−1)
第二种
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g(a,b,c,n)=\sum_{i=0}^ni\frac{ai+b}{c}
g(a,b,c,n)=i=0∑nicai+b
情况一：
a
≥
c
o
r
b
≥
c
a\ge c~or~b \ge c
a≥c or b≥c
g
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6
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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}
g(a,b,c,n)=g(a%c,b%c,c,n)+6n(n+1)(2n+1)⋅ca+2n(n+1)⋅cb
情况二：
a
<
b
a
n
d
a
<
c
a<b~and~a<c
a<b and a<c
g
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g(a,b,c,n)=\sum_{i=0}^ni\sum_{j=1}^m[ai+b\ge cj]=
g(a,b,c,n)=i=0∑nij=1∑m[ai+b≥cj]=
1
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\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)=
21(mn(n+1)−j=0∑m−1(acj+(c−b+a−1))2+j=0∑m−1acj+(c−b+a−1))=
1
2
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h
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\frac{1}{2}(mn(n+1)-h(c,c-b+a-1,a,m-1)+f(c,c-b+a-1,a,m-1))
21(mn(n+1)−h(c,c−b+a−1,a,m−1)+f(c,c−b+a−1,a,m−1))
第三种
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h(a,b,c,n)=\sum_{i=0}^n\left(\frac{ai+b}{c}\right)^2
h(a,b,c,n)=i=0∑n(cai+b)2
情况一：
a
≥
c
o
r
b
≥
c
a\ge c~or~b \ge c
a≥c or b≥c
h
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6
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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)+
h(a,b,c,n)=6n(n+1)(2n+1)⋅(ca)2+(cb)2⋅(n+1)+ca⋅cbn(n+1)+
2
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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)
2cag(a%c,b%c,c,n)+2cbf(a%c,b%c,c,n)+h(a%c,b%c,c,n)
情况二：
a
<
b
a
n
d
a
<
c
a<b~and~a<c
a<b and a<c
h
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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)=
h(a,b,c,n)=i=0∑nj=1∑m[ai+b≥cj](2j−1)=j=0∑m−1(2j+1)(n+1−acj+(c−b+a−1))=
m
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m^2(n+1)-2g(c,c-b+a-1,a,m-1)-f(c,c-b+a-1,a,m-1)
m2(n+1)−2g(c,c−b+a−1,a,m−1)−f(c,c−b+a−1,a,m−1)
大板子
#include <bits/stdc++.h>
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;
}
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