[ACM-ICPC 2018 沈阳网络赛] G. Spare Tire （思维+容斥)

Posted 2022-04-05 zhangbuang

A sequence of integer \lbrace a_n \rbracean? can be expressed as:

 

\displaystyle a_n = \left\ \beginarraylr 0, & n=0\\ 2, & n=1\\ \frac3a_n-1-a_n-22+n+1, & n>1 \endarray \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_ibi?is relatively-prime with the integer mm. And then calculate:

\displaystyle \sum_i=1^pa_b_ii=1∑p?abi??

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

SOLUTION:
考虑容斥，用所有的和减去不合法的和
也就是减去ai，其中i和m的gcd不为1
考虑对m进行质因子分解
对于m的每一质因子p我们需要减去小标为p的倍数的a
写出来之后发现可以推出来式子o（1）的进行计算，但是俩个质因子p可能
筛掉一个ai多次，加上容斥就行了

CODE:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll mod = 1e9+7;
 5 const ll inv6 = 166666668;
 6 const ll inv2 = 500000004;
 7 ll a[10005];
 8 ll clac(ll n,ll i)
 9     n /= i;
10     return (n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod * i % mod * i % mod + n % mod * (n + 1) % mod * inv2 % mod * i % mod) % mod;
11 
12 int main()
13     ll n,m;
14     while(~scanf("%lld%lld",&n,&m))
15         int cnt = 0;
16         for(int i = 2; i * i <= m; i++)
17             if(m % i == 0)
18                 a[cnt++] = i;
19                 while(m % i == 0)
20                     m /= i;
21             
22         
23         if(m != 1)
24             a[cnt++] = m;
25         ll ans = clac(n,1);
26         ll ans2 = 0;
27         for(int i = 1; i < (1 << cnt); i++)
28             int flag = 0;
29             ll tmp = 1;
30             for(int j = 0; j < cnt; j++)
31                 if(i & (1 << j))
32                     flag++;
33                     tmp = tmp * a[j] % mod;
34                 
35             
36             tmp = clac(n,tmp);
37             if(flag & 1) ans2 = (ans2 % mod + tmp % mod) % mod;
38             else ans2 = (ans2 % mod - tmp % mod + mod) % mod;
39         
40         printf("%lld\n",(ans % mod - ans2 % mod + mod) % mod);
41     
42     return 0;
43