NYOJ 题目998 Sum（欧拉函数，水）

来源：互联网发布：小学生题目解答软件编辑：程序博客网时间：2024/06/10 05:28

GCD

时间限制：1000 ms | 内存限制：65535 KB

难度：3

描述

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,please answer sum of X satisfies 1<=X<=N and (X,N)>=M.

输入

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 (1<=N<=10^9, 1<=M<=10^9), representing a test case.

输出

Output the answer mod 1000000007

样例输入

31 110 210000 72

样例输出

1351305000

上传者 ACM_张书军 丫的今天做题不顺，深夜水道水题压压惊 ac代码 #include<stdio.h>#include<string.h>long long eular(long long n){int i;long long ans=n;for(i=2;i*i<=n;i++){if(n%i==0){ans-=ans/i;while(n%i==0)n/=i;}}if(n>1)ans-=ans/n;return ans;}int main(){long long n,m;while(scanf("%lld%lld",&n,&m)!=EOF){int i;long long ans=0;for(i=1;i*i<=n;i++){if(n%i==0){if(i>=m)ans+=i*eular(n/i);if(i*i!=n&&n/i>=m)ans+=n/i*eular(i);}}printf("%lld\n",ans);}}

0 0