题解 | many sum

#include <iostream>

using namespace std;

const int N=2e6+5;

int n,m;
int a[N];
int b[N];

int main()
{
    scanf("%d%d%d",&n,&a[1],&m);
    for(int i=2;i<=n;i++) a[i] = (a[i-1]+7*i)%m;
    int ans=0;
    for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j+=i)
            b[j]+=a[i];
    for(int i=1;i<=n;i++) ans^=b[i];
    printf("%d",ans);
    return 0;
}