Question

Lazy Student Problem Description There is a test of Algorithms. Teacher provides a question bank consisting of N questions and guarantees all the questions in the test will be from this question bank. Due to lack of time and his laziness, Codu could only practice M questions. There are T questions in a question paper selected randomly. Passing criteria is solving at least 1 of the T problems. Codu can't solve the question he didn't practice. What is the probability that Codu will pass the test? Constraints 0 < T <= 10000 0 < N, T <= 1000 0 <= M <= 1000 M,T <= N Input Format First line contains single integer T denoting the number of test cases. First line of each test case contains 3 integers separated by space denoting N, T, and M. Output For each test case, print a single integer. If probability is p/q where p & q are co-prime, print (p*mulInv(q)) modulo 1000000007, where mulInv(x) is multiplicative inverse of x under modulo 1000000007. Timeout 1 Test Case Example 1 Input

Answer 1

Answer:

There is a test of Algorithms. Teacher provides a question bank consisting of N questions and guarantees all the questions in the test will be from this question bank. Due to lack of time and his laziness, Codu could only practice M questions. There are T questions in a question paper selected randomly. Passing criteria is solving at least 1 of the T problems. Codu can’t solve the question he didn’t practice. What is the probability that Codu will pass the test?

Input Format

First line contains single integer T denoting the number of test cases.

First line of each test case contains 3 integers separated by space denoting N, T, and M.

Output Format

For each test case, print a single integer.

If probability is p/q where p & q are co-prime, print (p*mulInv(q)) modulo 1000000007, where mulInv(x) is multiplicative inverse of x under modulo 1000000007.

Constraints

0 < T <= 10000

0 < N, T <= 1000

0 <= M <= 1000

M,T <= N

Sample Input and Output

Example 1

Input

1

4 2 1

Output

500000004

Solution —

#include< >

int combinations(int n, int r)

{

int itr,numerator=1,denominator=1,result;

for(itr=1; itr<=r; itr++)

{

denominator = denominator*itr;

numerator = numerator*(n-(itr-1));

}

result = numerator/denominator;

return result;

}

int calGCD(int num1,int num2)

{

int rem;

while (1)

{

rem = num1 % num2;

if(rem == 0)

return num2;

if(rem != 0)

{

num1 = num2;

num2 = rem;

}

}

}

int mulInv(int a)

{

unsigned long long int m=1000000007,itr,b;

for (itr = 1; itr 0)

{

scanf(“%llu\t%llu\t%llu”,&qb_questions,&choosen,&learnt);

unknown = qb_questions – learnt;

waysOfChoosing = combinations(qb_questions,choosen);

waysOfFailing = combinations(unknown,choosen);

p = waysOfChoosing – waysOfFailing;

q = waysOfChoosing;

gcd = calGCD(p,q);

if(gcd != 1)

{

p=p/gcd;

q=q/gcd;

}

result = (p*mulInv(q))%1000000007;

printf(“%llu”,result);

T–;

}

return 0;

}

Answer 2

Language used: Python 3.0

Program:

import math

n=int(input())

for i in range(n):

   N,T,M = map(int,input().split(' '))

   probability=(1-(math.factorial(N-M))/(math.factorial(T)*math.factorial(N-M)))

   #taking in the values of p,q from probability as of p/q

   p,q=probability.as_integer_ratio()

   #creating the multiple inverse of q

   print(pow(q,1000000007-2,1000000007))

Input:

1

4 2 1

Output:

500000004

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