BZOJ2956 (模积和)

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题目：2956: 模积和

题意：求∑∑((n mod i)*(m mod j))其中1<=i<=n,1<=j<=m,i≠j。然后mod 19940417

本题坑我太久啊，思路：

∑∑((n mod i) * (m mod j)) 1<=i<=n, 1<=j<=m, i≠j= ∑(n mod i) * ∑(m mod i) - ∑((n mod i) * (m mod i))= ∑(n-[n/i]*i) * ∑(m-[m/i]*i) - ∑(nm-([n/i]+[m/i])i+[n/i][m/i]*i*i)

#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const LL MOD = 19940417;LL sum(LL n){    return n*(n+1)%MOD*(2*n+1)%MOD*3323403%MOD;}LL Solve(LL m, LL n){    LL ans=0,i,last;    for(i=1;i<=m;i=last+1)    {        last=min(m,n/(n/i));        ans += (n/i)*(i+last)%MOD*(last-i+1)%MOD*9970209%MOD;        ans %= MOD;    }    return ans;}int main(){    LL n, m;    cin>>n>>m;    if(n<m) swap(n, m);    LL ans=(n*n-Solve(n,n))%MOD*((m*m-Solve(m,m))%MOD);    ans+=-m*m%MOD*n%MOD+Solve(m, n)*m%MOD+Solve(m, m)*n%MOD;    ans%=MOD;    for(LL i=1,last;i<=m;i=last+1)    {        last=min(m,min(n/(n/i),m/(m/i)));        ans += -(n/i)*(m/i)%MOD*((sum(last)-sum(i-1))%MOD)%MOD;        ans %= MOD;    }    cout<<(ans%MOD+MOD)%MOD<<endl;    return 0;}