Pku2992 (approx., prime factor decomposition)

Http: // 162.105.81.212/judgeonline/problem? Id = 2992

 

Question: calculate the number of comments C [N] [K. (0 <= k <= n <= 431)

Idea: the approximate number of a number num is CNT, And the num prime factor is decomposed to obtain num = p1 ^ A1 * P2 ^ A2 * P3 ^ A3 *...... * PN ^.

Then the number of approx. CNT = (A1 + 1) * (A2 + 1) * (A3 + 1 )*...... * (An + 1 ).

C [N] [k] = n! /(N-k )! * K !).

First, calculate the prime number table from 1 to 431. It is easy to time out without preprocessing.

1. Number of approx. Theorem: Let's divide the standard prime factor of N into n = p1 ^ A1 * P2 ^ A2 *... * PM ^ am,
Then the number of factors of N = (A1 + 1) * (A2 + 1) *... * (Am + 1 ).
2. n! The prime factor of = (n-1 )! Prime Factor + N prime factor
3. prime factor decomposition of C (n, k) = n! Prime factor-(n-k )! Prime factor-k! Prime Factor

# Include <iostream> <br/> using namespace STD; <br/> const int maxn = 432; <br/> int res [maxn] [maxn]; // n! <Br/> bool s [maxn]; <br/> int sumprime, prime [maxn]; <br/>/* int P [83] = {2, 3, 5, 7, 11, 13, 17,19, 23,29, 31,37, 41,43, 47,53, 59,61, 67,71, 83,89, 97, <br/> 101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197, <br/> 199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313, <br/> 317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431}; <Br/> name the table 250 ms */<br/> void prime () // evaluate the prime number 172 Ms. <br/>{< br/> int I, K, T; <br/> for (I = 2; I <= 435; I + = 2) {s [I] = 0; s [I-1] = 1 ;}< br/> S [1] = 0; s [2] = 1; <br/> for (I = 3; I <= 30; I + = 2) <br/> {<br/> If (s [I]) <br/> {<br/> K = 2 * I; t = I + k; <br/> while (T <= 435) {s [T] = 0; T = T + k ;}< br/>}< br/> K = 0; prime [0] = 2; <br/> for (I = 3; I <= 435; I + = 2) <br/> {<br/> If (s [I] = 1) {k ++; prime [k] = I ;}< br/>}< br/> sumprime = K; <Br/> // for (I = 0; I <sumprime; I ++) <br/> // printf ("% d", prime [I]); <br/>}< br/> void fun () // integer decomposition n! = P1 ^ T1 *... * PK ^ TK evaluate n! Prime factor decomposition of <br/>{< br/> int I, j, temp; <br/> for (I = 2; I <= 431; I ++) <br/> {<br/> // For (j = 0; j <= I; j ++) <br/> // res [I] [J] = res [I-1] [J]; <br/> temp = I; <br/> for (j = 0; j <sumprime; j ++) <br/>{< br/> while (temp> 1 & TEMP % prime [J] = 0) <br/> {<br/> res [I] [J] ++; // Number of prime factors + +; <br/> temp/= prime [J]; <br/>}< br/> for (I = 3; I <= 431; I ++) <br/> for (j = 0; j <sumprime; j ++) <br/> res [I] [J] + = res [I-1] [J]; <br/>}< br /> Int main () <br/>{< br/> int K, N, I; <br/>__ int64 ans; <br/> prime (); <br/> fun (); <br/> while (scanf ("% d", & N, & K )! = EOF) <br/>{< br/> ans = 1; <br/> for (I = 0; I <sumprime; I ++) // number of factors of N = (A1 + 1) * (A2 + 1 )*... * (Am + 1 ). <br/> ans * = (1 + (RES [N] [I]-res [k] [I]-res [n-k] [I]); <br/> printf ("% i64d/N", ANS); <br/>}< br/> return 0; <br/>}