【BZOJ】2956：模积和

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cmath>  
#include<algorithm>  
#include<iostream>  
#define mod 19940417   
using namespace std;  
long long ans,n,m,ine;  

long long power(long long x,long long y)  
{  
    long long ans=1;  
    while (y)  
    {  
        if (y&1) ans=ans*x%mod;  
        x=x*x%mod;  
        y>>=1;  
    }  
    return ans;  
}  
  
long long sum(long long x)  
{  
    return x*(x+1)/2%mod;  
}  
  
long long Sum(long long x)  
{  
    return x*(x+1)%mod*(2*x+1)%mod*ine%mod;  
}  
  
long long cal(long long n)  
{  
    long long ans=n*n%mod;  
    long long last;  
    for (long long i=1;i<=n;i=last+1)  
    {  
        last=n/(n/i);  
        ans=(ans-(n/i)*(sum(last)-sum(i-1)+mod)%mod+mod)%mod;  
    }  
    return ans%mod;  
}  
  
int main()  
{  
    scanf("%lld%lld",&n,&m);  
    ine=3323403;  
    if (n>m) swap(n,m);  
    ans=cal(n)*cal(m)%mod;  //计算式子1  [最好用long long，防止相乘时爆炸] 
    long long last;  
    for (long long i=1;i<=n;i=last+1)  
    {  
        last=min(n/(n/i),m/(m/i));//n/i=k,last=n/k  
        ans=(ans-n*m%mod*(last-i+1)%mod+m*(n/i)%mod*(sum(last)-sum(i-1)+mod)%mod+n*(m/i)%mod*(sum(last)-sum(i-1)+mod)%mod-(n/i)*(m/i)%mod*(Sum(last)-Sum(i-1)+mod)%mod+2*mod)%mod;  
    }  
    printf("%lld\n",ans);  
    return 0;  
}