HDU 2588 GCD [欧拉函数]【数论】
2016-08-11 20:25
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题目连接 : 传送阵
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GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1734 Accepted Submission(s): 852
Problem Description
The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.
Input
The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.
Output
For each test case,output the answer on a single line.
Sample Input
3
1 1
10 2
10000 72
Sample Output
1
6
260
—————————————–.
题目大意 :不用解释了吧 ..
解题思路:
这道题所需要的算法主要为欧拉函数的运用和一点点的GCD知识。
问题所要求的是 gcd( x , n ) > m ,由gcd( x , n )本身可知,gcd求出来的是 x 和n的最大公约数(设为a),即有式子gcd( x ,n )=a , 进一步进行化简可变为gcd( x/a , n/a )=1 , 到了此处这个式子又有了另一层含义——x/a与n/a互素 。在联想到欧拉函数的功能——对正整数n,欧拉函数是小于或等于n的数中与n互质的数的数目。于是将欧拉函数里的n换成n/a,不就正好能求出x/a的个数了吗?x/a的个数不就是我们所要求的x的个数了吗?
转自这里
附本题代码
————————————-.
————————-.
GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1734 Accepted Submission(s): 852
Problem Description
The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.
Input
The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.
Output
For each test case,output the answer on a single line.
Sample Input
3
1 1
10 2
10000 72
Sample Output
1
6
260
—————————————–.
题目大意 :不用解释了吧 ..
解题思路:
这道题所需要的算法主要为欧拉函数的运用和一点点的GCD知识。
问题所要求的是 gcd( x , n ) > m ,由gcd( x , n )本身可知,gcd求出来的是 x 和n的最大公约数(设为a),即有式子gcd( x ,n )=a , 进一步进行化简可变为gcd( x/a , n/a )=1 , 到了此处这个式子又有了另一层含义——x/a与n/a互素 。在联想到欧拉函数的功能——对正整数n,欧拉函数是小于或等于n的数中与n互质的数的数目。于是将欧拉函数里的n换成n/a,不就正好能求出x/a的个数了吗?x/a的个数不就是我们所要求的x的个数了吗?
转自这里
附本题代码
————————————-.
#include <stdio.h> #include <vector> #include <iostream> #include <stdlib.h> using namespace std; #define LL long long int #define pb push_back int Euler(int n) { if(n==1) return 1; int m=n; for(int i=2; i*i<=m; i++) if(m%i==0) { n-=n/i; while(m%i==0) m/=i; } if(m!=1) { n-=n/m; } return n; } int solve(int n,int m) { int ans=0; for(int i=1; i*i<=n; i++) { if(n%i) continue; if(i>=m&&i*i!=n) ans += Euler(n/i); if(n/i>=m) ans += Euler(i); } return ans; } int main() { int _,p=0; scanf("%d",&_); while(_--) { int n,m; scanf("%d%d",&n,&m); int sum=solve(n,m); printf("%d\n",sum); } return 0; }
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