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--- 2020-2021:teams:legal_string:王智彪:类欧几里得算法 [2021/08/16 10:49]
王智彪
+++ 2020-2021:teams:legal_string:王智彪:类欧几里得算法 [2021/08/17 11:32] (当前版本)
王智彪
@@ 行 86: / 行 86: @@
 $=g(a\ mod\ c,b\ mod\ c,c,n)+2 \lfloor {\frac a c} \rfloor h(a\ mod\ c,b\ mod\ c,c,n)+2 \lfloor {\frac b c} \rfloor f(a\ mod\ c,b\ mod\ c,c,n)+\sum_{i=0}^{n}{\lfloor {\frac a c} \rfloor}^{2} i^{2}+2 \lfloor {\frac a c} \rfloor \lfloor {\frac b c} \rfloor i+{\lfloor {\frac b c} \rfloor}^{2}$
+$=g(a\ mod\ c,b\ mod\ c,c,n)+2 \lfloor {\frac a c} \rfloor h(a\ mod\ c,b\ mod\ c,c,n)+2 \lfloor {\frac b c} \rfloor f(a\ mod\ c,b\ mod\ c,c,n)$
+$+{\frac {n(n+1)(2n+1)} 6}{\lfloor {\frac a c} \rfloor}^{2}+n(n+1)\lfloor {\frac a c} \rfloor \lfloor {\frac b c} \rfloor+(n+1){\lfloor {\frac b c} \rfloor}^{2}$
+$h(a,b,c,n)=\sum_{i=0}^{n}i\lfloor {\frac {i(a\ mod\ c)+b\ mod\ c} c} \rfloor+i(i\lfloor {\frac a c} \rfloor +\lfloor {\frac b c} \rfloor)$
+$=h(a\ mod\ c,b\ mod\ c,c,n)+{\frac {n(n+1)(2n+1)} 6}\lfloor {\frac a c} \rfloor+{\frac {n(n+1)} 2}\lfloor {\frac b c} \rfloor$
+当 $a＜c$ 且 $b＜c$ 时，仍设 $m=\lfloor {\frac {an+b} c} \rfloor$
+$g(a,b,c,n)=2\sum_{i=0}^{n}\sum_{j=1}^{\lfloor {\frac {ai+b} c} \rfloor}j-\sum_{i=0}^{n}\lfloor {\frac {ai+b} c} \rfloor$
+$=-f(a,b,c,n)+2\sum_{i=0}^{n}\sum_{j=1}^{m}j[j≤{\lfloor {\frac {ai+b} c} \rfloor}]$
+$=-f(a,b,c,n)+2\sum_{i=0}^{n}\sum_{j=1}^{m-1}(j+1)[(j+1)c＜ai+b+1]$
+$=-f(a,b,c,n)+2\sum_{j=0}^{m-1}(j+1)\sum_{i=0}^{n}[i>{\lfloor {\frac {jc+c-b-1} {a}} \rfloor}]$
+$=-f(a,b,c,n)+2\sum_{j=0}^{m-1}(j+1)(n-{\lfloor {\frac {jc+c-b-1} {a}} \rfloor})$
+$=nm(m+1)-f(a,b,c,n)-2h(c,-b-1,a,m)$
+$h(a,b,c,n)=\sum_{i=0}^{n}i\sum_{j=1}^{m}[j≤\lfloor {\frac {ai+b} {c}} \rfloor]$
+$=\sum_{i=0}^{n}i\sum_{j=0}^{m-1}[(j+1)c＜ai+b+1]$
+$=\sum_{j=0}^{m-1}\sum_{i=0}^{n}i[i＞{\frac {jc+c-b-1} {a}}]$
+$=\sum_{j=0}^{m-1}({\frac {n(n+1)} {2}}-\sum_{i=0}^{n}i[i≤{\lfloor {\frac {jc+c-b-1} {a}} \rfloor}])$
+$=\sum_{j=0}^{m-1}({\frac {n(n+1)} {2}}-{\frac {{\lfloor {\frac {jc+c-b-1} {a}} \rfloor}(\lfloor {\frac {jc+c-b-1} {a}} \rfloor+1)} {2}})$
+$={\frac {\sum_{j=0}^{m-1}n(n+1)-\sum_{j=0}^{m-1}{\lfloor {\frac {jc+c-b-1} {a}} \rfloor}^{2}-\sum_{j=0}^{m-1}{\lfloor {\frac {jc+c-b-1} {a}} \rfloor}} {2}}$
+$={\frac 1 2}[mn(n+1)-g(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1)]$
+注意负数那里会导致计算错误，比如 ${\frac {-1} 2}={\frac {1} 2}=0$
+所以在实际计算中，要规避掉参数为负的情况，这个具体情况具体分析。
+<hidden 代码>
+<code cpp>
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+const ll mod=998244353,inv2=(mod+1)>>1,inv3=332748118;
+struct Ans {
+	ll f,s,t;
+}ans,now,las;
+Ans getans(int a,int b,int c,int n) {
+	if(!a) {
+		now.f=(ll)b/c*(n+1)%mod;
+		now.s=(ll)(b/c)*(b/c)%mod*(n+1)%mod;
+		now.t=((ll)n+1)*n/2%mod*(b/c)%mod;
+	}
+	else if(a>=c||b>=c) {
+		las=getans(a%c,b%c,c,n);
+		now.f=(((ll)n*(n+1)/2%mod*(a/c)%mod+((ll)n+1)*(b/c)%mod)%mod+las.f)%mod;
+		ll tmp1=((las.s+2ll*(a/c)%mod*las.t%mod)%mod+2ll*(b/c)%mod*las.f%mod)%mod;
+		ll tmp2=(((ll)n*(n+1)%mod*(2ll*n+1)%mod*inv2%mod*inv3%mod*(a/c)%mod*(a/c)%mod+(ll)n*(n+1)%mod*(a/c)%mod*(b/c)%mod)%mod+(ll)(n+1)*(b/c)%mod*(b/c)%mod)%mod;
+		now.s=(tmp1+tmp2)%mod;
+		now.t=(((ll)n*(n+1)/2%mod*(b/c)%mod+((ll)n+1)*n%mod*inv2%mod*((ll)n*2+1)%mod*inv3%mod*(a/c)%mod)%mod+las.t)%mod;
+	}else {
+		ll m=((ll)a*n+b)/c;
+		las=getans(c,c-b-1,a,m-1);
+		now.f=((ll)n*m%mod-las.f+mod)%mod;
+		now.s=((ll)n*m%mod*(m+1)%mod-now.f+mod-2ll*las.t%mod+mod-2ll*las.f%mod+mod)%mod;
+		now.t=((ll)m*n%mod*(n+1)%mod-las.f+mod-las.s+mod)%mod*inv2%mod;
+	}
+	return now;
+}
+int n,a,b,c;
+int main() {
+	int t;
+	scanf("%d",&t);
+	while(t--) {
+		scanf("%d %d %d %d",&n,&a,&b,&c);
+		ans=getans(a,b,c,n);
+		printf("%lld %lld %lld\n",ans.f,ans.s,ans.t);
+		//printf("%lld %lld %lld\n",f(a,b,c,n),g(a,b,c,n),h(a,b,c,n));
+	}
+	return 0;
+}
+/*
+7 7 7
+0 10 4
+2 4 3
+4 5 3
+3 10 1
+0 0 7
+9 10 1
+5 5 4
+4 0 9
+4 4 9
+*/
+</code>
+</hidden>
+$ps$ ：这里一起计算是因为，单独记忆化 $t$ 了...

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