A sequence of integer \lbrace a_n \rbrace{an} can be expressed as:
\displaystyle a_n = \left\{ \begin{array}{lr} 0, & n=0\\ 2, & n=1\\ \frac{3a_{n-1}-a_{n-2}}{2}+n+1, & n>1 \end{array} \right.an=⎩⎨⎧0,2,23an−1−an−2+n+1,n=0n=1n>1
Now there are two integers nn and mm. I'm a pretty girl. I want to find all b_1,b_2,b_3\cdots b_pb1,b2,b3⋯bp that 1\leq b_i \leq n1≤bi≤n and b_ibiis relatively-prime with the integer mm. And then calculate:
\displaystyle \sum_{i=1}^{p}a_{b_i}i=1∑pabi
But I have no time to solve this problem because I am going to date my boyfriend soon. So can you help me?
Input
Input contains multiple test cases ( about 1500015000 ). Each case contains two integers nn and mm. 1\leq n,m \leq 10^81≤n,m≤108.
Output
For each test case, print the answer of my question(after mod 1,000,000,0071,000,000,007).
Hint
In the all integers from 11 to 44, 11 and 33 is relatively-prime with the integer 44. So the answer is a_1+a_3=14a1+a3=14.
样例输入复制
4 4
样例输出复制
14
如果要看详细的解释可以看这一篇，我觉得这个公式还是挺好推的。
https://blog.csdn.net/qq_39792342/article/details/82533693
之所以写这篇博客是因为要记住一个教训，在质数打表的时候需要开大一点的空间，比如说1e8的数据，就不能只开1e4的质数数组，不然一直TLE 或者WA。
下面是代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#define fori(l,r) for( int i = l ; i <= r ; i++ )
#define forj(l,r) for( int j = l ; j <= r ; j++ )
#define mem(a,val) memset(a,val,sizeof a)
#define inf 0x3f3f3f3f
#define longinf 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
ll n,m;
const int maxn = 1e5+5;
bool isprime[maxn];
ll prime[maxn];
ll a[maxn];
int cnt,cc;
const ll mod = 1e9+7;
ll six;
ll ans;
ll down;
const ll Six = 166666668;
const ll Two = 500000004; //乘法逆元用的
ll fastpow( ll base,ll x )
{
ll ans = 1;
while( x )
{
if( x&1 )
ans = (ans*base)%mod;
x = x>>1;
base = (base*base)%mod;
}
return ans;
}
void init()
{
cnt = 1;
mem(isprime,true);
for( int i = 4 ; i < maxn ; i+=2 )
isprime[i] = false;
for( int i = 3 ; i < maxn ; i+=2 )
if( isprime[i] )
for( int j = i<<1 ; j < maxn ; j+=i )
isprime[j] = false;
fori(2,maxn-1)
if( isprime[i] )
prime[cnt++] = i;
down = fastpow(2,mod-2);
six = fastpow(6,mod-2);
}
void decompose( ll x )
{
for( int i = 1 ; prime[i]*prime[i] <= x ; i++ )
{
if( x%prime[i] == 0 )
{
while( x%prime[i] == 0 )
x = x/prime[i];
a[cc++] = prime[i];
}
}
if( x > 1 )
a[cc++] = x;
}
inline ll Mod(ll a,ll b)
{
return (a%mod)*(b%mod)%mod;
}
inline ll ff(ll x)
{
ll k = n/x;
return (Mod(x,x)*Mod(Mod(k,k+1),Mod(k+k+1,Six))%mod + Mod(Mod(1+k,k),Mod(x,Two)) )%mod;
}
void dfs( int start,int times,int goal,ll val )
{
if( times == goal )
{
if( times&1 )
ans = ( ans+ff(val) )%mod;
else ans = ( ans-ff(val)+mod )%mod;
return;
}
for( int i = start ; i < cc ; i++ )
{
dfs(i+1,times+1,goal,val*a[i]);
}
}
int main()
{
init();
while( scanf("%lld %lld",&n,&m) == 2 )
{
ans = 0;
cc = 1;
decompose(m);
for( int i = 1 ; i < cc ; i++ )
dfs(1,0,i,1);
ans = ( ff(1)-ans+mod )%mod;
printf("%lld\n",ans);
}
return 0;
}
/*
99999281 99999827
100000000 100000000
9999999 9999999
9999998 9999998
9999997 9999997
9999998 9999997
100000000 9999997
6666666 6666666
6666666 6666667
*/