- 17.64%
- 1000ms
- 131072K
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
题目来源
题意：
已知一个数列a 和整数n， m
现在想知道1-n中所有与m互质的数 作为下标的a的和
思路：
预先打表处理的话内存不够 推公式能推出a[i] = （i + 1） * i【虽然我好像没有推出来...】
所以Sn也是可以推出来的
某一个数的倍数的a求和也是可以推的 因为是ki 那么提出一个k来 就可以代入公式求了
小于n 与m互质的数没有什么特殊的规律 但应该想到素数筛时的做法 用上容斥
先求出1-n所有的a的和 再减去所有与m不互质的数
用容斥来求不互质的数 是正是负与质因数的个数有关 如果是奇数个质因数之积的话就是加 偶数就是减
用到了乘法逆元的模板
cal（）函数注释掉的部分是WA的 改成了题解的方法就AC了
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
#include<cstring>
#include<set>
//#include<bits/stdc++.h>
#define inf 0x7f7f7f7f7f7f7f7f
using namespace std;
typedef long long LL; const int maxn = 1e5 + ;
const LL mod = 1e9 + ;
LL inv6, inv2;
LL n, m; LL p[maxn];
int cnt;
void getprime(LL x)
{
cnt = ;
for (LL i = ; i * i <= x; i++) {
if (x % i == ) {
p[cnt++] = i;
}
while (x % i == ) {
x /= i;
}
}
if (x > ) {
p[cnt++] = x;
}
} void ex_gcd(int a, LL b, LL &d, LL &x, LL &y)
{
if(!b){
d = a;
x = ;
y = ;
}
else{
ex_gcd(b, a % b, d, y, x);
y -= x * (a / b);
}
} int mod_inverse(int a, LL m)
{
LL x, y, d;
ex_gcd(a, m, d, x, y);
return (m + x % m) % m;
} LL cal(LL n, LL k)
{
/*LL ans = n / k * (n / k + 1) % mod;
ans = ans * (2 * n / k + 1) % mod;
ans = ans * inv6 % mod * k % mod * k % mod;
ans = ans + k * (1 + n / k) % mod * n / k % mod * inv2 % mod;*/
n=n/k;
return (n%mod*(n+)%mod*(*n+)%mod*inv6%mod*k%mod*k%mod+n%mod*(n+)%mod*inv2%mod*k%mod)%mod;
//return ans;
} int main()
{
inv6 = mod_inverse(, mod);
inv2 = mod_inverse(, mod);
//cout<<mod_inverse(6, mod)<<endl<<mod_inverse(2, mod)<<endl;
while (scanf("%lld%lld", &n, &m) != EOF) {
memset(p, , sizeof(p));
getprime(m);
LL ans = ;
for(int i = ; i < ( << cnt); i++){
int flag = ;
LL tmp = ;
for(int j = ; j < cnt; j++){
if(i & ( << j)){
flag++;
tmp = tmp * p[j] % mod;
} }
tmp = cal(n, tmp) % mod;
if(flag % ){
ans = (ans % mod + tmp % mod) % mod;
}
else{
ans = (ans % mod - tmp % mod + mod) % mod;
}
}
printf("%lld\n", (cal(n, ) % mod - ans % mod + mod) % mod);
}
return ;
}