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**Solution**

here the conventional process will fail when input is very large (i.e: N=2000000 & M=0)

so some tricks are used here. factors divisible by 5 or 2 ignore first.

i.e: 5*2*3=30 that means a pair of 2&5 produce trailing zero.

Henece all 2 or 5 facotrs are ignored. then I multiply the extra 5 or 2.

i.e:let there is two 5 & three 2. so here is two pairs of 2 & 5 and one extra 2.

Another tricks is applied: only last digit is enough in successive multiplication

to determine the last digit of final result.

i.e: 6*7*3=126. the last digist is 6. we can get the same result as follows:

1st step: 6*7=42(last digit is 2. now we will consider 42 as 2)

2nd step: 2*3=6

Thus this gives the same result. It is true for all cases.

this tricks is must for large input.

i.e: (i.e: N=2000000 & M=0)

in convention approach you may think as follows:

1st step: 2000000*199999*199998*…….2*1

2nd step: find out the last digit of the above multiplication result.

But there is no such big variable to store such large multiplication result.

**Source Code**

#include<stdio.h> int main() { int m,n,t,f,i,k,result,factorNotdivisibleBy5or2; while(scanf("%d%d",&n,&m)==2) { t=0; f=0; result=1; for(i=n; i>n-m; i--) { factorNotdivisibleBy5or2=i; while(!(factorNotdivisibleBy5or2&1)) { factorNotdivisibleBy5or2>>=1; t++; } while(factorNotdivisibleBy5or2%5==0) { factorNotdivisibleBy5or2/=5; f++; } result=(result*factorNotdivisibleBy5or2)%10; } for(i=t; i<f; i++) { result=(result*5)%10; } for(i=f; i<t; i++) { result=(result*2)%10; } printf("%d\n",result%10); } return 0; }

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