K. Sum of the Line

Consider a triangle of integers, denoted by TT. The value at (r,c)(r,c) is denoted by T_{r,c}Tr,c, where 1 \le r1≤r and 1 \le c \le r1≤c≤r. If the greatest common divisor of rr and cc is exactly 11, T_r,c = cTr,c=c, or 0 otherwise.

Now, we have another triangle of integers, denoted by SS. The value at (r,c)(r,c) is denoted by S_rSr,cc, where 1 \le r1≤r and 1 \le c \le r1≤c≤r. S_{r,c}Sr,c is defined as the summation \sum_{i=c}^{i} T_{r,i}∑i=ciTr,i.

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

Input

The first line of input contains an integer t (1 \le t \le 10000)t(1≤t≤10000) which is the number of test cases. Each test case includes a single line with an integer kk described as above satisfying 2 \le k \le 10^82≤k≤108.

Output

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

样例输入

2
2
3

样例输出

1
5

题意：

给你一个数字n，求

$\sum_{i=1}^ni^2(gcd(n,i)=1)$

思路：

可以说是非常水的容斥了，感觉只要把题意理解，知道是求上面那个式子就一定能过

对于n的所有质数，暴力枚举其倍数对答案的贡献即可，因为质数最多不超过9个，所以可以直接 $2^9$ 暴力

#include<stdio.h>
#define LL long long
#define mod 998244353
#define nv 166374059
LL A(LL x)  {  return x*(x+1)%mod*(2*x+1)%mod*nv%mod;  }
int pri[125];
int main(void)
{LL ans, now;int T, n, i, j, m, cnt, sum;scanf("%d", &T);while(T--){scanf("%d", &n);m = n, cnt = 0;for(i=2;i*i<=m;i++){if(m%i==0)pri[++cnt] = i;while(m%i==0)m /= i;}if(m!=1)pri[++cnt] = m;ans = A(n);for(i=1;i<(1<<cnt);i++){now = 1, sum = 0;for(j=0;j<=cnt-1;j++){if(i&(1<<j)){now *= pri[j+1];sum++;}}if(now<=n)ans = (ans-now*now%mod*A(n/now)*(sum%2?1:-1)%mod+mod)%mod;}printf("%lld\n", ans);}return 0;
}

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2017乌鲁木齐ICPC: K. Sum of the Line（容斥）

K. Sum of the Line

Input

Output

题意：

思路：

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