#P5493. 质数前缀统计

    ID: 6782 远端评测题 1000~3000ms 125MiB 尝试: 0 已通过: 0 显示难度省选/NOI− 上传者: 标签>数学素数判断,质数,筛法

质数前缀统计

Background

This is important prerequisite knowledge for the Zhouge sieve and the Min_25 sieve.

Problem Description

Let S(n)S(n) be the sum of the kk-th powers of all primes not exceeding nn.

Given NN, compute the value of the following expression:

$$\sum_{i=1}^{\lfloor\sqrt{N}\rfloor} i^2 S \!\left( \left\lfloor \frac{N}{i} \right\rfloor \right)$$

All results are taken modulo the given prime pp.

Input Format

One line contains three integers N,k,pN, k, p, separated by spaces.

Output Format

Output one integer on one line, representing the computed result.

10 3 1000000007
1458
100000 0 1000000007
941229402
100000 10 1000000007
446053671

Hint

Sample explanation:

$S \!\left( \left\lfloor \frac{N}{1} \right\rfloor \right)\! = S(10) = 2^3 + 3^3 + 5^3 + 7^3 = 503$.

$S \!\left( \left\lfloor \frac{N}{2} \right\rfloor \right)\! = S(5) = 2^3 + 3^3 + 5^3 = 160$.

$S \!\left( \left\lfloor \frac{N}{3} \right\rfloor \right)\! = S(3) = 2^3 + 3^3 = 35$.

$1^2 S \!\left( \left\lfloor \frac{N}{1} \right\rfloor \right)\! + 2^2 S \!\left( \left\lfloor \frac{N}{2} \right\rfloor \right)\! + 3^2 S \!\left( \left\lfloor \frac{N}{3} \right\rfloor \right)\! = 503 + 640 + 315 = 1458$.

Constraints

Test Point ID NN \le kk \le Time Limit
131\sim 3 10610^6 1010 1s1\texttt s
474\sim 7 4×10104\times {10}^{10} 00 3s3\texttt s
8128\sim 12 1010

For 100%100\% of the testdata, 0k100 \le k \le 10, 1N4×10101 \le N \le 4\times {10}^{10}, 109<p<1.01×109{10}^9 < p < 1.01 \times {10}^9.

Translated by ChatGPT 5