HDU 2588 GCD Euler + 暴力技巧
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任意门:http://acm.hdu.edu.cn/showproblem.php?pid=2588
GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3608 Accepted Submission(s): 1954
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.
(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
Source
题意概括:
求 1~N 的范围内存在多少个 X 使得 GCD( X, N ) >= M;
解题思路:
设 s = GCD( X, N);
可知: s >= M,
且存在 a, b 使得 s*a = X, s*b = N, GCD( a, b ) = 1;
因为 X <= N 所以 a <= b;
综上所述:
N 1e9 的范围缩小一半枚举 s ,求得 b;(因为可以同时求得 i 和 N/i 的方案数)
即求满足 GCD(a, b) = 1 且 a <= b 的 a 的个数。
AC code:
1 #include <bits/stdc++.h> 2 #define INF 0x3f3f3f3f 3 #define LL long long 4 using namespace std; 5 6 const int MAXN = 1e9+10; 7 LL N, M; 8 9 LL Euler(LL n) 10 { 11 LL res = n; 12 for(LL i = 2; i*i <= n; i++){ 13 if(n%i == 0) res = res/i*(i-1); 14 while(n%i == 0) n/=i; 15 } 16 if(n > 1) res = res/n*(n-1); 17 return res; 18 } 19 20 int main() 21 { 22 int T_case; 23 scanf("%d", &T_case); 24 LL ans, b; 25 while(T_case--){ 26 scanf("%lld %lld", &N, &M); 27 ans = 0; 28 for(LL s = 1; s*s <= N; s++){ 29 if(N%s) continue; 30 if(s >= M) ans+=Euler(N/s); 31 if(s*s != N && N/s >= M) ans+=Euler(s); 32 } 33 printf("%lld ", ans); 34 } 35 return 0; 36 }
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