C++ Exercises: Accepts n different numbers and s which is equal to the sum of the n different numbers
C++ Basic: Exercise-80 with Solution
Write a C++ program that accepts n different numbers (0 to 100) as well as an integer s which is equal to the sum of the n different numbers.
Your job is to find the number of combinations of n numbers and the same number cannot be used for one combination
Example:
Here n = 3 and s = 6:
1 + 2 + 3 = 6
0 + 1 + 5 = 6
0 + 2 + 4 = 6
Output: Number of combination: 3
Visual Presentation:
Sample Solution:
C++ Code :
#include <iostream>
// Macros for defining range loops
#define range(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define rep(i,n) range(i,0,n)
using namespace std;
int n, s;
long long int dp[10][1010]; // Declaration of 2D array to store combinations
int main(void) {
dp[0][0] = 1LL; // Initialize the first element of the array to 1
// Loop for calculating combinations
rep(i, 101) {
for (int j = 8; j >= 0; j--) {
rep(k, 1010) {
if (k + i <= 1010) {
dp[j + 1][k + i] += dp[j][k]; // Calculate combinations and store in the array
}
}
}
}
// Input values for 'n' and 's'
cout << "Input n and s: ";
cin >> n >> s;
// Output the number of combinations for given 'n' and 's'
cout << "\nNumber of combinations: ";
cout << dp[n][s] << endl;
return 0;
}
Sample Output:
Input n and s: 3 6 Number of combination: 3
Flowchart:
C++ Code Editor:
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