P2260 [清华集训2012]模积和

$Hyperlink$

https://www.luogu.com.cn/problem/P2260

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$Description$

P2834换个模数，此时模数不是质数

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$Solution$

用$exgcd$或欧拉定理求逆元即可

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$Code$

#include
#include
#include
#define LL long long
#define mod 19940417
using namespace std;LL n,m,inv2,inv6,S1,S2,Ans;
inline LL read()
{
     
	char c;LL d=1,f=0;
	while(c=getchar(),!isdigit(c)) if(c=='-') d=-1;f=(f<<3)+(f<<1)+c-48;
	while(c=getchar(),isdigit(c)) f=(f<<3)+(f<<1)+c-48;
	return d*f;
}
inline LL sum1(LL x){
     return x*(x+1)%mod*inv2%mod;}
inline LL sum2(LL x){
     return x*(x+1)%mod*(2*x+1)%mod*inv6%mod;}
signed main()
{
     
	n=read();m=read();
	if(n>m) n^=m,m=n^m,n^=m;
	inv2=9970209;inv6=3323403;
	S1=n*n%mod;
	for(register int l=1,r;l<=n;l=r+1)
	{
     
		r=n/(n/l);
		S1-=(sum1(r)-sum1(l-1)+mod)%mod*(n/l)%mod;
		S1=(S1+mod)%mod;
	}
	S2=m*m%mod;S2=(S2+mod)%mod;
	for(register int l=1,r;l<=m;l=r+1)
	{
     
		r=m/(m/l);
		S2-=(sum1(r)-sum1(l-1)+mod)%mod*(m/l)%mod;
		S2=(S2+mod)%mod;
	}
	Ans=S1*S2%mod;
	Ans-=n*n%mod*m%mod;Ans=(Ans+mod)%mod;
	for(register int l=1,r;l<=n;l=r+1)
	{
     
		r=min(n/(n/l),m/(m/l));
		Ans+=(sum1(r)-sum1(l-1)+mod)%mod*(m*(n/l)%mod+n*(m/l)%mod)%mod;Ans=(Ans+mod)%mod;
		Ans-=(sum2(r)-sum2(l-1)+mod)%mod*(m/l)%mod*(n/l)%mod;Ans=(Ans+mod)%mod;
	}
	printf("%lld",Ans);
}