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Topic: 100! (Read 946 times) |
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sat90
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Posts: 7
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whats the logic behind calculating 100!......w/o using any bigint lib
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SMQ
wu::riddles Moderator
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Re: 100!
« Reply #1 on: Aug 15th, 2011, 9:19am » |
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Since 100! has 158 (decimal) digits, you're going to be hard-pressed to calculate the exact result without an arbitrary-precision math library of some sort. However you can, for instance, count the number of zeros at the end w/o a bigint library.
--SMQ
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--SMQ
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: 100!
« Reply #2 on: Aug 15th, 2011, 1:14pm » |
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It's not that hard to implement the method of multiplication you learned in school. Although it's not very efficient, it'll get the job done.
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Wikipedia, Google, Mathworld, Integer sequence DB
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Grimbal
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Re: 100!
« Reply #3 on: Aug 22nd, 2011, 5:07am » |
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Or you can apply the method you learned at school. With a pencil, paper and lots of time.
100! should still be feasible.
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sat90
Newbie
Posts: 7
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Re: 100!
« Reply #5 on: Sep 5th, 2011, 6:10am » |
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i m not able to simulate manual multiplication using algorithm...plz help!!
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towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
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Re: 100!
« Reply #6 on: Sep 5th, 2011, 8:49am » |
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Something like:
Pseudo Code:
int A[200] = {0};
A[0]=1;
for(C=2; i <=100; i++)
{
temp = 0;
for(i=0; i < 200; i++)
{
temp += C * a[i];
A[i] = temp % 10;
temp /= 10;
}
}
for(i=200; i > 0 && A[i] == 0; i--)
;
for(; i >=0; i--)
cout << a[i];
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