## Algorithm

n common cubic dice are thrown. What is the probability that the sum of all thrown dice is at least x? Input The input file contains several test cases. Each test case consists two integers n (1 ≤ n ≤ 24) and x (0 ≤ x < 150). The meanings of n and x are given in the problem statement. Input is terminated by a case where n =0 and x =0. This case should not be processed. Output For each line of input produce one line of output giving the requested probability as a proper fraction in lowest terms in the format shown in the sample output. All numbers appearing in output are representable in unsigned 64-bit integers. The last line of input contains two zeros and it should not be processed.

Sample Input 3 9 1 7 24 24 15 76 24 56 24 143 23 81 7 38 0 0

Sample Output 20/27 0 1 11703055/78364164096 789532654692658645/789730223053602816 25/4738381338321616896 1/2 55/46656

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````#include <bits/stdc++.h>

using namespace std;
#define ll long long

vector < vector<ll>> memo(25, vector < ll>(151,-1));

ll solve(int n, int x) {
if(n == 0)
return x == 0 ? 1 : 0;
if(memo[n][x] != -1) return memo[n][x];
ll res = 0;
for(int i=1;i<=6;i++) {
// can't go below 0, already meet requirement
int nextX = max(x - i, 0);
res += solve(n-1, nextX);
}
memo[n][x] = res;
return res;
}

ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}

int main()
{
int n,x;
while(scanf("%d %d",&n,&x),(n+x)) {
ll numerator = solve(n,x);
ll denominator = floor(pow(6, n));
if(numerator == denominator) {
printf("1\n");
} else if(numerator != 0) {
ll d = gcd(numerator, denominator);
numerator /= d;
denominator /= d;
printf("%lld/%lld\n", numerator, denominator);
} else {
printf("0\n">;
}
}
}
``````
Copy The Code &

Input

cmd
3 9
1 7
24 24
15 76
24 56
24 143
23 81
7 38
0 0

Output

cmd
20/27
0
1
11703055/78364164096
789532654692658645/789730223053602816
25/4738381338321616896
1/2
55/