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hdu2824 The Euler function(欧拉函数个数)
转载请注明出处:http://blog.csdn.net/u012860063
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2824
欧拉函数性质:
1:(百科):http://baike.baidu.com/link?url=r-yneKCCyS9N6bhbQCqiZX0V2OCYq9r7iHSzHTSs03H7qRvu1OfUzlOxfVEs2PmR
2:http://www.cppblog.com/doer-xee/archive/2009/12/01/102353.html
Problem Description
The 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)
Input
There 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
Source
2009 Multi-University Training Contest 1 - Host by TJU
题意:求a到b中每个数的欧拉函数的总和!
代码如下:
#include <iostream> #include <cstdio> #include <cmath> using namespace std; #define N 3000001 __int64 phi[N]; void init() { int i, j; for(i = 1; i < N; i++) phi[i] = i; for(i = 2; i < N; i++) if(i == phi[i]) //此时i为素数 for(j = i; j < N; j += i) //j累加i phi[j] = (phi[j] / i) * (i - 1); //j有因子i,而且i是素数,正是欧拉函数 } int main() { init(); int a, b; while(scanf("%d%d", &a, &b) != EOF) { __int64 ans = 0; for(int i = a; i <= b; i++) ans += phi[i]; printf("%I64d\n", ans); } return 0; }
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