ICPC沈阳预赛 C题 Convex Hull

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/Myriad_Dreamin/article/details/82533698

定义一个函数

$$s_2(x)=s^2(x)=\left\{\begin{aligned}0&,x=ar^2,\forall r\in N_+\\x&\text{,otherwise}\end{aligned}\right.$$
问 $\displaystyle \sum_{i=1}^n\sum_{j=1}^is_2(j)$的值.
运用交换主元的方法：
$\displaystyle\begin{aligned} \sum_{i=1}^n\sum_{j=1}^is_2(j)=&\sum_i\sum_j[1\leqslant j\leqslant i][1\leqslant i\leqslant n]s_2(j)\\=&\sum_i\sum_j[1\leqslant j\leqslant i\leqslant n]s_2(j)\\=&\sum_i\sum_j[1\leqslant j\leqslant n][j\leqslant i\leqslant n]s_2(j)\\=&\sum_{j=1}^ns_2(j)\sum_{i=j}^n1=\sum_{j=1}^n(n-j+1)s_2(j)\\=&(n+1)\sum_{j=1}^n s^2(j)-\sum_{j=1}^n s^3(j)\end{aligned}$
运用容斥就可以求出两项和式，单次复杂度由下式给出
$$T(n)=\mathrm{O}(\frac{\sqrt{n}}{\log n})+\sum_{i=2}^{\frac{\sqrt{n}}{\log n}}T(\frac{n}{i^2})$$
不会解，但是很容易知道这个式子是不太超过 $O(\sqrt{n})$量级的。
$\text{PS:}T(n)\approx \mathrm{O}(\frac{\sqrt{n}}{\log n})+T(\frac{6n}{\pi^2})\approx \mathrm{O}(\sqrt{n})$
啊啊啊，怎么这么简单！考试没做真可惜了。

#include <cstdio>
const int N=100005,lgN=20;
int np[N+5],prim[N/(lgN-5)],ptop=0;
void sieve(){
    for(int i=2;i<=N;i++){
        if(np[i])continue;
        prim[ptop++]=i;
        for(int j=i+i;j<=N;j+=i)np[j]=1;
    }
}
long long n,p;
inline long long mul(long long x,long long y){
    long long tmp=(x*y-(long long)((long double)x/p*y+1.0e-8)*p);
    return tmp<0?tmp+p:tmp;
}
inline long long cal2(long long n){
    long long n2=n+1,n3=2*n+1;
    if((n&1)==0)n>>=1;else n2>>=1;
    if((n%3==0))n/=3;else if(n2%3==0)n2/=3;else n3/=3;
    return mul(n,mul(n2,n3));
}
inline long long cal3(long long n){
    long long n2=n,n3=n+1,n4=n+1;
    if((n&1)==0)n>>=1,n2>>=1;
    else n3>>=1,n4>>=1;
    return mul(mul(n,n2),mul(n3,n4));
}
long long inc2(long long x,int dep){
    long long res=0,pri=(long long)prim[dep]*prim[dep],t;
    for(int i=dep;pri<=x;i++,pri=(long long)prim[i]*prim[i]){
        t=mul(pri,pri);
        res+=mul(t,cal2(x/pri)-inc2(x/pri,i+1)+p);
    }
    return res;
}
long long inc3(long long x,int dep){
    long long res=0,pri=(long long)prim[dep]*prim[dep],t;
    for(int i=dep;pri<=x;i++,pri=(long long)prim[i]*prim[i]){
        t=mul(pri,mul(pri,pri));
        res+=mul(t,cal3(x/pri)-inc3(x/pri,i+1)+p);
    }
    return res;
}

int main(){
    sieve();
    while(~scanf("%lld%lld",&n,&p)){
        long long res=mul(n+1,(cal2(n)-inc2(n,0)+p)%p)-(cal3(n)-inc3(n,0))%p+p;
        printf("%lld\n",res%p);
    }
}