- 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_ibi?is relatively-prime with the integer mm. And then calculate:
\displaystyle \sum_{i=1}^{p}a_{b_i}i=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
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
1 #include<iostream>
2 #include<stdio.h>
3 #include<string.h>
4 #include<algorithm>
5 #include<stack>
6 #include<queue>
7 #include<map>
8 #include<vector>
9 #include<cmath>
10 #include<cstring>
11 #include<set>
12 //#include<bits/stdc++.h>
13 #define inf 0x7f7f7f7f7f7f7f7f
14 using namespace std;
15 typedef long long LL;
16
17 const int maxn = 1e5 + 5;
18 const LL mod = 1e9 + 7;
19 LL inv6, inv2;
20 LL n, m;
21
22
23 LL p[maxn];
24 int cnt;
25 void getprime(LL x)
26 {
27 cnt = 0;
28 for (LL i = 2; i * i <= x; i++) {
29 if (x % i == 0) {
30 p[cnt++] = i;
31 }
32 while (x % i == 0) {
33 x /= i;
34 }
35 }
36 if (x > 1) {
37 p[cnt++] = x;
38 }
39 }
40
41 void ex_gcd(int a, LL b, LL &d, LL &x, LL &y)
42 {
43 if(!b){
44 d = a;
45 x = 1;
46 y = 0;
47 }
48 else{
49 ex_gcd(b, a % b, d, y, x);
50 y -= x * (a / b);
51 }
52 }
53
54 int mod_inverse(int a, LL m)
55 {
56 LL x, y, d;
57 ex_gcd(a, m, d, x, y);
58 return (m + x % m) % m;
59 }
60
61 LL cal(LL n, LL k)
62 {
63 /*LL ans = n / k * (n / k + 1) % mod;
64 ans = ans * (2 * n / k + 1) % mod;
65 ans = ans * inv6 % mod * k % mod * k % mod;
66 ans = ans + k * (1 + n / k) % mod * n / k % mod * inv2 % mod;*/
67 n=n/k;
68 return (n%mod*(n+1)%mod*(2*n+1)%mod*inv6%mod*k%mod*k%mod+n%mod*(n+1)%mod*inv2%mod*k%mod)%mod;
69 //return ans;
70 }
71
72
73 int main()
74 {
75 inv6 = mod_inverse(6, mod);
76 inv2 = mod_inverse(2, mod);
77 //cout<<mod_inverse(6, mod)<<endl<<mod_inverse(2, mod)<<endl;
78 while (scanf("%lld%lld", &n, &m) != EOF) {
79 memset(p, 0, sizeof(p));
80 getprime(m);
81 LL ans = 0;
82 for(int i = 1; i < (1 << cnt); i++){
83 int flag = 0;
84 LL tmp = 1;
85 for(int j = 0; j < cnt; j++){
86 if(i & (1 << j)){
87 flag++;
88 tmp = tmp * p[j] % mod;
89 }
90
91 }
92 tmp = cal(n, tmp) % mod;
93 if(flag % 2){
94 ans = (ans % mod + tmp % mod) % mod;
95 }
96 else{
97 ans = (ans % mod - tmp % mod + mod) % mod;
98 }
99 }
100 printf("%lld\n", (cal(n, 1) % mod - ans % mod + mod) % mod);
101 }
102 return 0;
103 }
原文地址：https://www.cnblogs.com/wyboooo/p/9646334.html