• 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
    所有与k不互质的数的贡献就是p1的倍数的贡献+p2的倍数的贡献+...+pu的倍数的贡献-p1*p2的倍数的贡献-p1*p3的倍数的贡献-...+p1*p2*p3的倍数的贡献+......
    
    以p1的倍数的贡献为例,他的贡献是(p1*1)^2+(p1*2)^2+...+(p1*[k/p1])^2,就是p^2*(1^2+2^2+...+(p1*[k/p1])^2
    
    
    #include <bits/stdc++.h>
    #define ll long long
    using namespace std;
    const int N=1e4+10;
    const int p=998244353;
    int T,cnt,k;
    bool vis[N];
    int prime[N],a[N];
    void pre()
    {
        for (int i=2;i<N;i++)
        {
            if (!vis[i]) prime[++cnt]=i;
            for (int j=1;j<=cnt&&prime[j]*i<N;j++)
            {
                vis[prime[j]*i]=1;
                if (i%prime[j]==0) break;
            }
    
        }
    }
    ll poww(ll x,int y)
    {
        ll ret=1;
        while (y)
        {
            if (y&1) ret=ret*x%p;
            x=x*x%p;
            y>>=1;
        }
        return ret;
    }
    int fund(int n)
    {
        int sum=0;
        for (int i=1;i<=cnt;i++)
        {
            if (prime[i]>n) break;
    
            if (n%prime[i]==0)
            {
                a[sum++]=prime[i];
                while (n%prime[i]==0) n/=prime[i];
            }
        }
        if (n>1) a[sum++]=n;
        return sum;
    }
    void solve(int n)
    {
        ll nn=(ll)n;
        ll inv6=poww(6,p-2);
        ll ans=nn%p*(nn+1)%p*(2*nn+1)%p*inv6%p;
    
        int sum=fund(n);
    
        ll tmp=0;
        for (int i=1;i<(1<<sum);i++)
        {
            ll x=1;int s=0;
            for (int j=0;j<sum;j++)
            {
                if ((i>>j)&1)
                {
                    x=x*(ll)a[j]%p;
                    s++;
                }
            }
    
            ll t=(ll)n/x;
            if (s&1) tmp=(tmp+x*x%p*t%p*(t+1)%p*(2*t+1)%p*inv6%p)%p;
            else tmp=((tmp-x*x%p*t%p*(t+1)%p*(2*t+1)%p*inv6%p)%p+p)%p;
    
        }
    
        ans=((ans-tmp)%p+p)%p;
    
        printf("%lld
    ",ans);
    }
    int main()
    {
        pre();
        scanf("%d",&T);
        while (T--)
        {
            scanf("%d",&k);
            solve(k);
        }
        return 0;
    }
    View Code
     
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  • 原文地址:https://www.cnblogs.com/tetew/p/9513904.html
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