Search found 2 matches

by Kamarul Kawnayeen
Thu Feb 03, 2011 12:39 pm
Forum: Volume 117 (11700-11799)
Topic: 11743 - Credit Check
Replies: 6
Views: 3485

Re: 11743 - Credit Check Why WA please help me!!!!!

If you can change your approaches to input the card number, then the problem become easy. Try the following code segment for inputing card number: int credit[16], j; for(j=0;j<16;j++) { scanf("%1d", &credit[j]); } Then sum the number in odd index of the array credit[] (i.e. at 1, 3, 5,...15 index) a...
by Kamarul Kawnayeen
Wed Feb 02, 2011 10:26 pm
Forum: Volume 2 (200-299)
Topic: 294 - Divisors
Replies: 91
Views: 23869

Re: 294 please help me

A better approach to solve this problem should be using prime divisor. If you can find that a number N = (a^x)*(b^y)*(c^z) where a, b, c are prime number then the number of divisor of N should be (x+1)*(y+1)*(Z+1). If you have already solve 583 (prime factor) then it will be easier for you to solve ...

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