Sum of the Line(ICPC2017 Urumqi)

题目描述

Consider a triangle of integers, denoted by T. The value at (r, c) is denoted by Tr,c , where 1 ≤ r and 1 ≤ c ≤ r. If the greatest common divisor of r and c is exactly 1, Tr,c = c, or 0 otherwise.
Now, we have another triangle of integers, denoted by S. The value at (r, c) is denoted by S r,c , where 1 ≤ r and 1 ≤ c ≤ r. S r,c is deﬁned as the summation

Here comes your turn. For given positive integer k, you need to calculate the summation of elements in k-th row of the triangle S.

输入

The first line of input contains an integer t (1 ≤ t ≤ 10000) which is the number of test cases.
Each test case includes a single line with an integer k described as above satisfying 2 ≤ k ≤ 10^8 .

输出

For each case, calculate the summation of elements in the k-th row of S, and output the remainder when it divided
by 998244353.

样例输入

2
2
3

样例输出

1
5


题意：对于gcd(i,k) == 1,则S(i,k) = i,否则为0，求S = sigma(S(i,k)^2)

题解：
如果没有gcd(i,k)这个限制，则答案就为signa(i^2) 即sigma(i^2) = n*(n+1)*(2*n+1)/6
gcd(i,k) != 1，说明i 与 k 存在共同的因子（除1外），对于n(<=1e8)的最多的素因子不超过10个，对因子进行容斥
可以表示为：|U| - |A∪B∪C∪....∪Z| ，A ，B ，...., Z 分别表示为能被某个素数整除的

首先将k中素因子筛选出来，用唯一分解定理，然后对这些因子进行容斥，可以dfs,也可以用二进制 （0,1表示取或不取，但必须保证至少取一次）

AC code:

#include <iostream>
#include <cstdio>
using namespace std;
const int MAXN = 10000000;
const int mod = 998244353;
typedef long long ll;

ll fast_pow(ll a,ll n)
{
    ll ans = 1;
    while(n)
    {
        if(n&1) ans = (ans * a)%mod;
        a = (a * a)%mod;
        n >>= 1;
    }
    return ans;
}
const ll six = fast_pow(6,mod-2);

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int tot = 0;
        ll n,k;
        scanf("%lld",&k);
        n = k;
        int prime[100];
        for(int i = 2;i * i <= n;i++)
        {
            if(n % i == 0)
            {
                prime[tot++] = i;
                while(n % i == 0)   n /= i;
            }
        }
        if(n != 1) prime[tot++] = n;
        ll ans = k*(k+1)%mod*(2*k+1)%mod*six%mod;
        for(int i = 1;i < (1<<tot);i++)
        {
            n = i;
            int pos = tot - 1,cnt = 0;
            ll ant = 1;
            while(n)
            {
                if(n&1) ant *= prime[pos],cnt++;
                pos--;
                n >>= 1;
            }
            ll m = k/ant;
            if(cnt&1)
                ans = (ans - ant*ant%mod*m%mod*(m + 1)%mod*(2*m + 1)%mod*six%mod + mod)%mod;
            else
                ans = (ans + ant*ant%mod*m%mod*(m + 1)%mod*(2*m + 1)%mod*six%mod)%mod;
        }
        printf("%lld
",ans);
    }
}