hdu 5608 function

function
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 478 Accepted Submission(s): 174

Problem Description

There is a function f(x),which is defined on the natural numbers set N,satisfies the following eqaution

N2−3N+2=∑d|Nf(d)

calulate ∑Ni=1f(i) mod 109+7.

Input

the first line contains a positive integer T,means the number of the test cases.

next T lines there is a number N

T≤500,N≤109

only 5 test cases has N>106.

Output

Tlines,each line contains a number,means the answer to the i-th test case.

Sample Input

1
3

Sample Output

2

1 2 −3∗1+2=f(1)=0 
2 2 −3∗2+2=f(2)+f(1)=0−>f(2)=0 
3 2 −3∗3+2=f(3)+f(1)=2−>f(3)=2 
f(1)+f(2)+f(3)=2 

Source

BestCoder Round #68 (div.2)

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【分析】
自己明明(假装)学过杜教筛，却被一道裸题卡了好久= =
…它就是裸题吧
注意前1e6的数据自己O(nlogn)的筛一遍先，要不TLE

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【代码】

//hdu#include<map>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define N 1000000#define ll long long#define inv2 500000004#define inv6 166666668#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int mod=1e9+7;const int mxn=1000005;map <int,int> M;int m,n,T;long long tmp[mxn];inline void init(){    int i,j;    fo(i,1,N) tmp[i]=(ll)(i-1)*(i-2)%mod;    fo(i,1,N)      for(j=i+i;j<=N;j+=i)        tmp[j]=(tmp[j]-tmp[i]+mod)%mod;    fo(i,1,N) tmp[i]=(tmp[i]+tmp[i-1])%mod;}inline int solve(int n){    if(n<=N) return tmp[n];    if(M.count(n)) return M[n];    ll sum=(ll)n*(n+1)%mod*(n+n+1)%mod*inv6%mod;    sum=(sum-(ll)3*n%mod*(n+1)%mod*inv2%mod)%mod;    sum=(sum+n+n)%mod;    for(int i=2,last=0;i<=n;i=last+1)    {        last=n/(n/i);        sum=(sum-(ll)(last-i+1)*solve(n/i)%mod)%mod;    }    sum=(sum+mod)%mod;    return M[n]=(int)sum;}int main(){    int i,j,T;    init();    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        printf("%d\n",solve(n));    }    return 0;}