INTA-004
# interview assessment questions
Dice Rolls
// C++ function to calculate the number of
// ways to achieve sum x in n no of throws
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
int dp[55][55];
// Function to calculate recursively the
// number of ways to get sum in given
// throws and [1..m] values
int NoofWays(int face, int throws, int sum)
{
// Base condition 1
if (sum == 0 && throws == 0)
return 1;
// Base condition 2
if (sum < 0 || throws == 0)
return 0;
// If value already calculated dont
// move into re-computation
if (dp[throws][sum] != -1)
return dp[throws][sum];
int ans = 0;
for (int i = 1; i <= face; i++) {
// Recusively moving for sum-i in
// throws-1 no of throws left
ans += NoofWays(face, throws - 1, sum - i);
}
// Inserting present values in dp
return dp[throws][sum] = ans;
}
// Driver function
int main()
{
int faces = 6, throws = 3, sum = 12;
memset(dp, -1, sizeof dp);
cout << NoofWays(faces, throws, sum) << endl;
return 0;
}
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You have d dice, and each die has f faces numbered 1, 2, ..., f.
Return the number of possible ways (out of fd total ways)
modulo 10^9 + 7 to roll the dice so the
sum of the face up numbers equals target.
Example 1:
Input: d = 1, f = 6, target = 3
Output: 1
Explanation:
You throw one die with 6 faces. There is only one way to get a sum of 3.
Example 2:
Input: d = 2, f = 6, target = 7
Output: 6
Explanation:
You throw two dice, each with 6 faces. There are 6 ways to get a sum of 7:
1+6, 2+5, 3+4, 4+3, 5+2, 6+1.
Example 3:
Input: d = 2, f = 5, target = 10
Output: 1
Explanation:
You throw two dice, each with 5 faces. There is only one way to get a sum of 10: 5+5.
Example 4:
Input: d = 1, f = 2, target = 3
Output: 0
Explanation:
You throw one die with 2 faces. There is no way to get a sum of 3.
Example 5:
Input: d = 30, f = 30, target = 500
Output: 222616187
Explanation:
The answer must be returned modulo 10^9 + 7.
Constraints:
1 <= d, f <= 30
1 <= target <= 1000
