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hdu 3501 Calculation 2

2024-11-21 05:18:02   204人阅读

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4551    Accepted Submission(s): 1879


Problem Description
Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.
 

 

Input
For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.
 

 

Output
For each test case, you should print the sum module 1000000007 in a line.
 

 

Sample Input
340
 

 

Sample Output
02
 

 

Author
GTmac
 

 

Source
2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT
 

 

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题意：给定N 求小于N 且与N不互素的数的和
欧拉函数性质 ： 与x互素的数的和=x*φ(x)/2
屠龙宝刀点击就送
#include <ctype.h>#include <cstdio>void read(long long &x){    x=0;bool f=0;register char ch=getchar();    for(;!isdigit(ch);ch=getchar()) if(ch==‘-‘) f=1;    for(; isdigit(ch);ch=getchar()) x=(x<<3)+(x<<1)+ch-‘0‘;    x=f?(~x)+1:x;}long long a=1,k;long long getphi(long long n){    long long ans=n;    if(n%2==0)    {        while(n%2==0) n/=2;        ans/=2;    }    for(int i=3;i*i<=n;i+=2)    {        if(n%i==0)        {            while(n%i==0) n/=i;            ans=ans/i*(i-1);        }    }    if(n>1) ans=ans/n*(n-1);    return ans;}int main(){    for(;a;)    {        read(k);        if(!k) break;        long long sum=(long long)k*(k-1)/2;sum-=(long long)getphi(k)*k/2;        printf("%lld\n",sum% 1000000007);//再忘了取模就剁手！！    }    return 0;}

 

hdu 3501 Calculation 2


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