Problem:
Score : 500 points

Problem Statement

Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total.

When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %.
Find the expected number of games that will be played, and print it as follows.

We can represent the expected value as P/Q with coprime integers P and Q.
Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}.
(Such an integer R always uniquely exists under the constraints of this problem.)

Constraints

1 \leq N \leq 100000

0 \leq A,B,C \leq 100

1 \leq A+B

A+B+C=100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C

Output

Print the expected number of games that will be played, in the manner specified in the statement.

Sample Input 1

1 25 25 50

Sample Output 1

2

Since N=1, they will repeat the game until one of them wins.
The expected number of games played is 2.

Sample Input 2

4 50 50 0

Sample Output 2

312500008

C may be 0.

Sample Input 3

1 100 0 0

Sample Output 3

1

B may also be 0.

Sample Input 4

100000 31 41 28

Sample Output 4

104136146

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <math.h>

int combination(int n,int r,int q);

int main(void){
    int i,j,k;
    int N = 0,A,B,C;
    double answerA = 0,answerB = 0;
    
    scanf("%d %d %d %d",&N,&A,&B,&C);
    
    double a = (double)A / 100,b = (double)B / 100,c = (double)C / 100;
    
    for(i = N;i < 300;i++){
        for(j = 0;j <= i - N && j < N;j++){
            k = i - N - j;
            answerA += i * pow(a,N - 1) * pow(b,j) * pow(c,k) * combination(N - 1,j,k) * a;
        }
    }
    
    for(i = N;i < 300;i++){
        for(j = 0;j <= i - N && j < N;j++){
            k = i - N - j;
            answerB += i * pow(b,N - 1) * pow(a,j) * pow(c,k) * combination(N - 1,j,k) * b;
        }
    }
    
    printf("%lf\n",answerA + answerB);
    
    return 0;
}

int combination(int n,int r,int q){
    int N = n + r + q;
    int a = 1,b = 1,c = 1,d = 1;
    for(a = 1;n > 0;n--) a = a * n;
    for(b = 1;r > 0;r--) b = b * r;
    for(c = 1;q > 0;q--) c = c * q;
    
    for(;N > 0;N--) d = d * N;
    
    return (d)/(a * b * c);
}