(Hdu step 7.2.1) The Euler function (Euler function template question -- calculate The sum of phi [a] to phi [B]), hdu7.2.1

(Hdu step 7.2.1) The Euler function (Euler function template question -- calculate The sum of phi [a] to phi [B]), hdu7.2.1

Question:

The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission (s): 166 Accepted Submission (s): 96

Problem DescriptionThe Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. here comes a very easy question: suppose you are given a, B, try to calculate (a) + (a + 1) + .... + (B)

InputThere are several test cases. Each line has two integers a, B (2 <a <B <3000000 ).

Output Output the result of (a) + (a + 1) +... + (B)

Sample Input 3 100

Sample Output 3042

Source2009 Multi-University Training Contest 1-Host by TJU

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Question Analysis:

Euler's function, simple question. Attackers can solve this problem directly .... The following describes some knowledge about Euler's functions.

1. Definition: for a positive integer n, Phi (n) is a positive integer smaller than or equal to n, and the number of mutual quality with n.

For example, Phi (8) = 4, because 1, 3, 5, 7 and 8 are mutually qualitative.

2. properties: 1) If p is a prime number, Phi (p) = p-1.

2) if n is the k power of prime p, Phi (n) = (p-1) * p ^ (k-1 ). Because all the multiples of p are mutually dependent on n.

3) the Euler's function is a product function. If m, n are mutually qualitative, Phi (mn) = PHI (m) PHI (n ).

Based on these three properties, we can introduce an integer Euler's function formula. Because a number can always be written as the product of some prime numbers.

E (k) = (p1-1) (p2-1 )... (pi-1) * (p1 ^ (a1-1) (p2 ^ (a2-1 ))... (pi ^ (ai-1 ))

= K * (p1-1) (p2-1)... (pi-1)/(p1 * p2 *... * pi)

= K * (1-1/p1) * (1-1/p2)... (1-1/pk)

Using the following functions in the program, you can quickly find the value of the Euler function (a is a qualitative factor of N)

If (N % a = 0 & (N/a) % a = 0), E (N) = E (N/a) *;

If (N % a = 0 & (N/a) %! = 0) then there are: E (N) = E (N/a) * (A-1 );

The Code is as follows:

/** A1.cpp ** Created on: March 19, 2015 * Author: Administrator */# include <iostream> # include <cstdio> using namespace std; const int maxn = 3000001; int phi [maxn];/*** initializes the Euler's array. * phi [8]: indicates from 1 ~ Number of 8-plus elements **/void prepare () {int I; for (I = 1; I <maxn; ++ I) {phi [I] = I;} int j; for (I = 2; I <maxn; ++ I) {if (phi [I] = I) {for (j = I; j <maxn; j + = I) {phi [j] = phi [j]/I * (I-1 );}}}} int main () {prepare (); int a, B; while (scanf ("% d", & a, & B )! = EOF) {long ans = 0; int I; for (I = a; I <= B; ++ I) {// obtain the sum between phi [a] and phi [B] and ans + = phi [I];} printf ("% lld \ n", ans );} return 0 ;}

The following is a version of TLE:

The only reason for the TLE is that every time you calculate phi [I], it drops phi (). The workload is too large.

/** POJ_2407.cpp ** Created on: July 15, November 19, 2013 * Author: Administrator */# include <iostream> # include <cstdio> # include <cstring> using namespace std; typedef long ll; const int maxn = 1000015; bool u [maxn]; ll su [maxn]; ll num; ll gcd (ll a, ll B) {if (B = 0) {return a;} return gcd (B, a % B);} void prepare () {// generate the prime number table ll I, j; memset (u, true, true, sizeof (u); for (I = 2; I <= 1000010; ++ I) {if (u [I]) {su [++ nu M] = I ;}for (j = 1; j <= num; ++ j) {if (I * su [j]> 1000010) {break ;} u [I * su [j] = false; if (I % su [j] = 0) {break ;}}} ll phi (ll x) {// Euler's function, used to calculate the number of integers in the [1, x) that are mutually correlated with x ll ans = 1; int I, j, k; for (I = 1; I <= num; ++ I) {if (x % su [I] = 0) {j = 0; while (x % su [I] = 0) {++ j; x/= su [I];} for (k = 1; k <j; ++ k) {ans = ans * su [I] % 10000%7ll;} ans = ans * (su [I]-1) % 10000%7ll; if (x = 1) {break ;}} I F (x> 1) {ans = ans * (x-1) % lower limit 7ll;} return ans;} int main () {prepare (); long; long B; while (scanf ("% lld", & a, & B )! = EOF) {long ans = 0; long I; for (I = a; I <= B; ++ I) {ans + = phi (I );} printf ("% lld \ n", ans);} return 0 ;}