f(N) = 00 + 11 + 22 + 33 + 44 + ... + NN 求模m的和。N很大，1e9级别，m不是很大，也就几十万。限时1s.
先利用取模的性质：(i + m * k) % m = i % m。对f(N)进行等价划分：
i ^ i + (i + m) ^ (i +m) + (i +2*m) ^ (i + 2*m) + ... = i ^ i + i ^ (i + m) + i ^ (i + 2*m) + ...
等式右边提出一个 a1 = i ^ i. 令 q = i ^ m。得 a1 * ( 1 + q + q^2 + q^3 + ...)
这虽然就成了等比数列，但是等比数列求和公式有分母，无法进行取模运算。即使求分母的逆元，但是若分母和m不互质，分母是没有逆元的(if the denominator and m are not coprime).
这时就需要发现技巧了。
1 + a + a^2 + ... + a^(2*n+1) = (1 + a) * (1 + (a^2) + (a^2)^2 + ... + (a^2)^n),
若最高次幂为偶数，那一项单独计算；否则，就用这个公式进行反复平方法（类似于快速幂的思想）在O(logn)时间内计算出一个几何级数的和。最多有m个这样的几何级数（等价划分的来的）求和。
note that 0^0 = 1.
实现见代码中的gp函数。
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
using namespace std;
#define ll long long
ll pow_mod(ll b, ll p, ll m) {
ll res = 1;
while (p) {
if (p & 1) res = res * b % m;
b = b * b % m;
p >>= 1;
}
return res;
}
//若用公式，必须用逆元解决除数取模的问题// (q^(n+1) - 1) / (q - 1) % m，但不幸的是……
ll gp(ll q, ll n, ll m) { // 1 + q + q^2 + ... + q^n
if (q == 0) return 1 % m;
ll factor = 1, sum = 0, temp;
while (n > 0) {
if (n % 2 == 0) {
temp = (factor * pow_mod(q, n, m)) % m; //偶数个和，先计算出幂指数最大的右边那项
sum = (sum + temp) % m;
n--;
}
factor = (((1 + q) % m) * factor) % m;
q = (q * q) % m; //仍然是反复平方法
n >>= 1; //底数平方一下，指数减半
}
return (sum + factor) % m;
}
int main()
{
ios::sync_with_stdio(0);
ll t;
cin >> t;
while (t--) {
ll n, m, s;
s = 0;
cin >> n >> m;
if (n < m)
for (ll i = 0; i <= n; i++)
s = (s + pow_mod(i, i, m)) % m;
else {
s = 1; // 0^0 = 1
for (ll i = 1; i < m; i++) {
ll cnt = (n - i) / m; //项数 - 1(a1)
ll a1 = pow_mod(i, i, m); //a1
ll q = pow_mod(i, m, m);
if (q == 1) //小心公比为1
s = (s + a1 * (cnt + 1)) % m;
else s = (s + ((a1* gp(q, cnt, m)) % m)) % m;
}
}
cout << s << endl;
}
return 0;
}
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Suhana and Equation Problem Code: SEQUASubmit
\
All submissions for this problem are available.
Suhana is studying following equation :
f(N) = 00 + 11 + 22 + 33 + 44 + ... + NN
Can you solve f(N) % M before Suhana does? \
Input
First line contains T, number of test cases.
First line of each test case contains N and M. \
Output
Output the required answer in new line.
Constraints
- 1 ≤ T ≤ 100
- 1 ≤ N ≤ 10^9
- 1 ≤ M ≤ 10^3
Example
Input:
4
1 9
2 29
3 93
40 993
Output:
2
6
33
907
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