Diophantus of Alexandria
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2269 Accepted Submission(s): 851
Problem Description
Diophantus
of Alexandria was an egypt mathematician living in Alexandria. He was
one of the first mathematicians to study equations where variables were
restricted to integral values. In honor of him, these equations are
commonly called diophantine equations. One of the most famous
diophantine equation is x^n + y^n = z^n. Fermat suggested that for n
> 2, there are no solutions with positive integral values for x, y
and z. A proof of this theorem (called Fermat's last theorem) was found
only recently by Andrew Wiles.
of Alexandria was an egypt mathematician living in Alexandria. He was
one of the first mathematicians to study equations where variables were
restricted to integral values. In honor of him, these equations are
commonly called diophantine equations. One of the most famous
diophantine equation is x^n + y^n = z^n. Fermat suggested that for n
> 2, there are no solutions with positive integral values for x, y
and z. A proof of this theorem (called Fermat's last theorem) was found
only recently by Andrew Wiles.
Consider the following diophantine equation:
1 / x + 1 / y = 1 / n where x, y, n ∈ N+ (1)
Diophantus
is interested in the following question: for a given n, how many
distinct solutions (i. e., solutions satisfying x ≤ y) does equation (1)
have? For example, for n = 4, there are exactly three distinct
solutions:
1 / 5 + 1 / 20 = 1 / 4
1 / 6 + 1 / 12 = 1 / 4
1 / 8 + 1 / 8 = 1 / 4
Clearly,
enumerating these solutions can become tedious for bigger values of n.
Can you help Diophantus compute the number of distinct solutions for big
values of n quickly?
Input
The
first line contains the number of scenarios. Each scenario consists of
one line containing a single number n (1 ≤ n ≤ 10^9).
first line contains the number of scenarios. Each scenario consists of
one line containing a single number n (1 ≤ n ≤ 10^9).
Output
The
output for every scenario begins with a line containing "Scenario #i:",
where i is the number of the scenario starting at 1. Next, print a
single line with the number of distinct solutions of equation (1) for
the given value of n. Terminate each scenario with a blank line.
output for every scenario begins with a line containing "Scenario #i:",
where i is the number of the scenario starting at 1. Next, print a
single line with the number of distinct solutions of equation (1) for
the given value of n. Terminate each scenario with a blank line.
Sample Input
2
4
1260
Sample Output
Scenario #1:
3
Scenario #2:
113
Source
Recommend
这题数论求的是 数的因子个数，设数为n，其可表示为
n=p1^r1 * p2^r2 * . . . * pn^rn
其中，p为素数，且可知其n的因子个数
k=(r1+1)*(r2+2)*...*(rn+1);
又由题可得，
1/x+y/1=1/n ==>
x>n && y>n ==>
xy=nx+ny，设y=n+k，x=n*(n+k)/k，即所求为n*n的因子个数
k=(2*r1+1)*(2*r2+2)*...*(2*rn+1);
注意结果要求多少对，故ans=k/2+1;
//140MS 200K 622 B G++
#include<stdio.h>
#include<math.h>
long long solve(int n)
{
long long ans=;
int i;
int m=(int)sqrt(n+0.5);
for(i=;i<=m;i++){
int ret=;
if(n%i==){
n/=i;
while(n%i==){
n/=i;ret++;
}
ans*=(*ret+);
}
if(n<i) break;
}
if(n>) ans*=;
return ans;
}
int main(void)
{
int n;
int cas=,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("Scenario #%d:\n%lld\n",cas++,solve(n)/+);
printf("\n");
}
return ;
}