这道题总是超时！好不容易 不超时了！当数在小的范围内是对的！但数字一达就错了！
Visible Lattice Points--------------------------------------------------------------------------------

Time limit: 1sec. Submitted: 73
Memory limit: 64M Accepted: 37

Source : Greater NewYork 2006

--------------------------------------------------------------------------------

A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points (x, y) with 0 <= x, y <= 5 with lines from the origin to the visible points.

Write a program which, given a value for the size, N, computes the number of visible points (x,y) with 0 <= x, y <= N.

Input

The first line of input contains a single integer C, (1 <= C <= 1000) which is the number of datasets that follow.

Each dataset consists of a single line of input containing a single integer N, (1 <= N <= 1000), which is the size.

Output

For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.

Sample Input

4
2
4
5
231
Sample Output

1 2 5
2 4 13
3 5 21
4 231 32549

这是我写的！
#include <stdio.h>

int main()
{
int m, n, i;
long a(int n);

scanf("%d", &m);
for(i = 1;i <= m;i ++)
{
scanf("%d", &n);
if(n == 1)
printf("%d %d %d\n", i, n, 3);
else if(n == 2)
printf("%d %d %d\n" ,i, n, 5);
else
printf("%d %d %ld\n", i, n, a(n));
}

return 0;
}

long a(int n)
{
long sum;
int b(int n);
if(n == 3)
sum = 9;
else
sum = b(n - 3) + a(n - 1);

return sum;
}

int b(int n)
{
int f1 = 0, f2 = 4, f;
int i;

for(i = 0;i < n;i ++)
{
f = f1 + f2;
f1 = f2;
f2 = f;
}

return f;
}

还有能帮我回答一下9楼的问题吗？谢谢了！

又写了一个，好像用时更多了！
#include <stdio.h>

int main()
{
long s, m, i;
long a(long n);

scanf("%ld", &s);
for(i = 1;i <= s;i ++)
{
scanf("%ld", &m);
if(m == 1)
printf("%ld %ld %d\n", i, m, 3);
else
printf("%ld %ld %ld\n", i, m, a(m));
}

return 0;
}

long a(long n)
{
long sum = 0;

if(n == 1)
sum = 3;
else if(n == 2)
sum = 5;
else if(n == 3)
sum = 9;
else
sum = a(n - 1) + a(n - 2) - 1;
return sum;
}

又写了一个，还是不对啊！
#include <stdio.h>

int main()
{
int s, m, i, j, k;
int sum;

scanf("%d", &s);
for(i = 1;i <= s;i ++)
{
sum = 0;
scanf("%d", &m);
if(m == 1)
printf("%d %d %d\n", i, m, 3);
else if(m == 2)
printf("%d %d %d\n", i, m, 5);
else if(m == 3)
printf("%d %d %d\n", i, m, 9);
else
{
for(k = 4;k <= m;k ++)
for(j = 2;j <= i - 2;j ++)
{
if(i%j == 0)
continue;
else if(i%2 == 0&&j%2 == 0)
continue;
else
sum ++;
}
sum += (3 + 2*(2*m -3));
printf("%d %d %d\n", i, m, sum);
}
}

return 0;

}