Count a * b
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 211 Accepted Submission(s): 116
Problem Description
Marry likes to count the number of ways to choose two non-negative integers a and b less than m to make a×b mod m≠0.
Let's denote f(m) as the number of ways to choose two non-negative integers a and b less than m to make a×b mod m≠0.
She has calculated a lot of f(m) for different m, and now she is interested in another function g(n)=∑m|nf(m). For example, g(6)=f(1)+f(2)+f(3)+f(6)=0+1+4+21=26. She needs you to double check the answer.
Give you n. Your task is to find g(n) modulo 264.
Input
The first line contains an integer T indicating the total number of test cases. Each test case is a line with a positive integer n.
1≤T≤20000
1≤n≤109
Output
For each test case, print one integer s, representing g(n) modulo 264.
Sample Input
2
6
514
6
514
Sample Output
26
328194
328194
Source
题意：略
分析：http://blog.csdn.net/firstlucker/article/details/49336427
这篇题解不错。
下面说明:x的约数的欧拉函数的和 = x
即sigma(d|x, phi(d)) = x
因为对于所有1-x的数，与x的gcd必定为x的约数，设为d，那这样的数有多少？phi(x / d)个
又发现x/d也是x的约数，所以。。。
sigma(d|x, phi(d)) = sigma(d|x, phi(x/d)) = x
这篇题解最后一步用了约数和定理。
其代码中防溢出运算更是简单、机智的令人发指
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <map>
#include <set>
#include <algorithm>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define MLL (1000000000000000001LL)
#define INF (1000000001)
#define For(i, s, t) for(int i = (s); i <= (t); i ++)
#define Ford(i, s, t) for(int i = (s); i >= (t); i --)
#define Rep(i, n) for(int i = (0); i < (n); i ++)
#define Repn(i, n) for(int i = (n)-1; i >= (0); i --)
#define mk make_pair
#define ft first
#define sd second
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define sz(x) ((int) (x).size())
#define clr(x, y) (memset(x, y, sizeof(x)))
inline void SetIO(string Name)
{
string Input = Name + ".in";
string Output = Name + ".out";
freopen(Input.c_str(), "r", stdin);
freopen(Output.c_str(), "w", stdout);
} inline int Getint()
{
char ch = ' ';
int Ret = ;
bool Flag = ;
while(!(ch >= '' && ch <= ''))
{
if(ch == '-') Flag ^= ;
ch = getchar();
}
while(ch >= '' && ch <= '')
{
Ret = Ret * + ch - '';
ch = getchar();
}
return Ret;
} const int N = ;
int n;
int Prime[N], Tot;
bool Visit[N]; inline void GetPrime()
{
For(i, , N-)
{
if(!Visit[i]) Prime[++Tot] = i;
For(j, , Tot)
{
if(i * Prime[j] >= N) break;
Visit[i * Prime[j]] = ;
if(!(i % Prime[j])) break;
}
}
} inline void Solve(); inline void Input()
{
GetPrime();
int TestNumber = Getint();
while(TestNumber--)
{
n = Getint();
Solve();
}
} inline void Solve()
{
if(n == )
{
puts("");
return;
} LL Total = , Except = n;
For(i, , Tot)
{
if(Prime[i] * Prime[i] > n) break;
if(!(n % Prime[i]))
{
int Fact = ;
LL Cnt = ;
while(!(n % Prime[i]))
{
Cnt *= Prime[i];
Fact++;
n /= Prime[i];
}
Except *= Fact;
Cnt *= Prime[i];
LL a = (Cnt - ) / (Prime[i] - ), b = Cnt + , c = Prime[i] + ;
Total *= ((a / c) * (b / c) * c + a % c * (b / c) + b % c * (a / c));
//cout << Total << ' ' << Except << endl;
}
} if(n > ) Except <<= , Total *= ( + 1LL * n * n);
cout << Total - Except << endl;
} int main()
{
SetIO("");
Input();
//Solve();
return ;
}