给定两个自然数n和m 。任务是找到可以将大于或等于m的数字相加以获得总和n的方式的数目。
例子：
Input : n = 3, m = 1
Output : 3
Following are three different ways
to get sum n such that each term is
greater than or equal to m
1 + 1 + 1, 1 + 2, 3
Input : n = 2, m = 1
Output : 2
Two ways are 1 + 1 and 2
这个想法是通过定义2D矩阵(例如dp [] [])来使用动态规划。 dp [i] [j]定义使用大于或等于j的数字获得和i的方式的数目。因此dp [i] [j]可以定义为：
If i < j, dp[i][j] = 0, because we cannot achieve smaller sum of i using numbers greater than or equal to j.
If i = j, dp[i][j] = 1, because there is only one way to show sum i using number i which is equal to j.
Else dp[i][j] = dp[i][j+1] + dp[i-j][j], because obtaining a sum i using numbers greater than or equal to j is equal to the sum of obtaining a sum of i using numbers greater than or equal to j+1 and obtaining the sum of i-j using numbers greater than or equal to j.
以下是此方法的实现：
C++
// CPP Program to find number of ways to
// which numbers that are greater than
// given number can be added to get sum.
#include
#define MAX 100
using namespace std;
// Return number of ways to which numbers
// that are greater than given number can
// be added to get sum.
int numberofways(int n, int m)
{
int dp[n+2][n+2];
memset(dp, 0, sizeof(dp));
dp[0][n + 1] = 1;
// Filling the table. k is for numbers
// greater than or equal that are allowed.
for (int k = n; k >= m; k--) {
// i is for sum
for (int i = 0; i <= n; i++) {
// initializing dp[i][k] to number
// ways to get sum using numbers
// greater than or equal k+1
dp[i][k] = dp[i][k + 1];
// if i > k
if (i - k >= 0)
dp[i][k] = (dp[i][k] + dp[i - k][k]);
}
}
return dp[n][m];
}
// Driver Program
int main()
{
int n = 3, m = 1;
cout << numberofways(n, m) << endl;
return 0;
} Java
// Java Program to find number of ways to
// which numbers that are greater than
// given number can be added to get sum.
import java.io.*;
class GFG {
// Return number of ways to which numbers
// that are greater than given number can
// be added to get sum.
static int numberofways(int n, int m)
{
int dp[][]=new int[n+2][n+2];
dp[0][n + 1] = 1;
// Filling the table. k is for numbers
// greater than or equal that are allowed.
for (int k = n; k >= m; k--) {
// i is for sum
for (int i = 0; i <= n; i++) {
// initializing dp[i][k] to number
// ways to get sum using numbers
// greater than or equal k+1
dp[i][k] = dp[i][k + 1];
// if i > k
if (i - k >= 0)
dp[i][k] = (dp[i][k] + dp[i - k][k]);
}
}
return dp[n][m];
}
// Driver Program
public static void main(String args[])
{
int n = 3, m = 1;
System.out.println(numberofways(n, m));
}
}
/*This code is contributed by Nikita tiwari.*/Python3
# Python3 Program to find number of ways to
# which numbers that are greater than
# given number can be added to get sum.
MAX = 100
import numpy as np
# Return number of ways to which numbers
# that are greater than given number can
# be added to get sum.
def numberofways(n, m) :
dp = np.zeros((n + 2, n + 2))
dp[0][n + 1] = 1
# Filling the table. k is for numbers
# greater than or equal that are allowed.
for k in range(n, m - 1, -1) :
# i is for sum
for i in range(n + 1) :
# initializing dp[i][k] to number
# ways to get sum using numbers
# greater than or equal k+1
dp[i][k] = dp[i][k + 1]
# if i > k
if (i - k >= 0) :
dp[i][k] = (dp[i][k] + dp[i - k][k])
return dp[n][m]
# Driver Code
if __name__ == "__main__" :
n, m = 3, 1
print(numberofways(n, m))
# This code is contributed by RyugaC#
// C# program to find number of ways to
// which numbers that are greater than
// given number can be added to get sum.
using System;
class GFG {
// Return number of ways to which numbers
// that are greater than given number can
// be added to get sum.
static int numberofways(int n, int m)
{
int[, ] dp = new int[n + 2, n + 2];
dp[0, n + 1] = 1;
// Filling the table. k is for numbers
// greater than or equal that are allowed.
for (int k = n; k >= m; k--) {
// i is for sum
for (int i = 0; i <= n; i++) {
// initializing dp[i][k] to number
// ways to get sum using numbers
// greater than or equal k+1
dp[i, k] = dp[i, k + 1];
// if i > k
if (i - k >= 0)
dp[i, k] = (dp[i, k] + dp[i - k, k]);
}
}
return dp[n, m];
}
// Driver Program
public static void Main()
{
int n = 3, m = 1;
Console.WriteLine(numberofways(n, m));
}
}
/*This code is contributed by vt_m.*/PHP
= $m; $k--)
{
// i is for sum
for ($i = 0; $i <= $n; $i++)
{
// initializing dp[i][k] to number
// ways to get sum using numbers
// greater than or equal k+1
$dp[$i][$k] = $dp[$i][$k + 1];
// if i > k
if ($i - $k >= 0)
$dp[$i][$k] = ($dp[$i][$k] + $dp[$i - $k][$k]);
}
}
return $dp[$n][$m];
}
// Driver Program
$n = 3;
$m = 1;
echo numberofways($n, $m) ;
return 0;
// This code is contributed by ChitraNayal
?>输出：
3