BZOJ 2956 模积和

题目链接：模积和

　　这种有模运算的题一般都要把取模运算给去掉，改成除法取下整的模式。即(amod b=a-lfloor frac{a}{b}  floor b)

　　然后我们先把(i=j)的也统计进答案，最后再减去即可。接下来就是推式子时间((n le m))：

egin{aligned}
&sum_{i=1}^nsum_{j=1}^m(n mod i)(m mod j) \
=&sum_{i=1}^n(n mod i)sum_{j=1}^m(m mod j) \
=&sum_{i=1}^n(n-ilfloor frac{n}{i} floor)sum_{j=1}^m(m-jlfloor frac{m}{j} floor) \
=&(n^2-sum_{i=1}^nilfloor frac{n}{i} floor)(m^2-sum_{i=1}^milfloor frac{m}{i} floor)
end{aligned}

egin{aligned}
&sum_{i=1}^n(n mod i)(m mod i) \
=&sum_{i=1}^n(n-ilfloor frac{n}{i} floor)(m-ilfloor frac{m}{i} floor) \
=&n^2m-msum_{i=1}^nilfloor frac{n}{i} floor-nsum_{i=1}^nilfloor frac{m}{i} floor+sum_{i=1}^nlfloor frac{n}{i} floorlfloor frac{m}{i} floor i^2
end{aligned}

　　然后分块计算即可。上下两个式子一减就是答案。

　　下面贴代码：

#include<cstdio>
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
#define mod 19940417
#define llg long long
#define min(a,b) (a<b?a:b)

const int n6=7776;
int n,m,a[3];

llg S(int x){return (1ll*x*(x+1)>>1)%mod;}
llg S2(int x){
	a[0]=x,a[1]=x+1,a[2]=2*x+1;
	for(int i=0;i<3;i++) if(a[i]%2==0){a[i]/=2;break;}
	for(int i=0;i<3;i++) if(a[i]%3==0){a[i]/=3;break;}
	return 1ll*a[0]*a[1]%mod*a[2]%mod;
}

llg work(int x,int y){
	llg now=0;
	for(int i=1,nt;i<=x;i=nt+1)
		nt=min(y/(y/i),x),now+=1ll*(y/i)*(S(nt)-S(i-1));
	return now%mod;
}

int main(){
	File("a");
	scanf("%d %d",&n,&m);
	if(n>m) n^=m^=n^=m;
	llg s1=work(n,n),ans;
	ans=(1ll*n*n-s1)%mod*(1ll*m*m%mod-work(m,m))%mod;
	ans+=s1*m+work(n,m)*n; ans%=mod;
	ans-=1ll*n*n%mod*m; ans%=mod;
	for(int i=1,nt;i<=n;i=nt+1){
		nt=min(n/(n/i),m/(m/i));
		ans-=(S2(nt)-S2(i-1))*(m/i)%mod*(n/i);
		ans%=mod;
	}
	printf("%lld",(ans+mod)%mod);
	return 0;
}