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
#include
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
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