给定一个N行M列的二维数组和一个整数K 。任务是找出是否有可能从给定数组的每一行中取出一个元素，并使总和等于 K。
例子：
Input: N = 2, M = 10, K = 5
arr = {{4, 0, 15, 3, 2, 20, 10, 1, 5, 4},
{4, 0, 10, 3, 2, 25, 4, 1, 5, 4}}
Output: YES
Explanation:
Take 2 from first row and 3 from second row.
2 + 3 = 5
So, we can make 5 by taking exactly one element
from each row.
Input:N = 3, M = 5, K = 5
arr = {{4, 3, 4, 5, 4},
{2, 2, 3, 4, 3},
{2, 1, 3, 3, 2}}
Output: NO方法：这个问题可以用动态规划解决。
- 我们可以制作一个 N 行 K 列的二维二进制数组 DP[][]。其中 DP[i][j] = 1 表示我们可以通过从每一行中取出一个元素直到 i 来使总和等于 j。
- 因此，我们将从 i = [0, N], k = [0, K] 迭代数组并检查当前 sum(k) 是否存在。
- 如果当前总和存在，那么我们将遍历该列并为每个可能小于或等于 K 的总和更新数组。
下面是上述方法的实现
C++
// C++ implementation to find
// whether is it possible to
// make sum equal to K
#include
using namespace std;
// Function that prints whether is it
// possible to make sum equal to K
void PossibleSum(int n, int m,
vector > v,
int k)
{
int dp[n + 1][k + 1] = { 0 };
// Base case
dp[0][0] = 1;
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= k; j++)
{
// Condition if we can make
// sum equal to current column
// by using above rows
if (dp[i][j] == 1)
{
// Iterate through current
// column and check whether
// we can make sum less than
// or equal to k
for (int d = 0; d < m; d++)
{
if ((j + v[i][d]) <= k)
{
dp[i + 1][j + v[i][d]] = 1;
}
}
}
}
}
// Printing whether is it
// possible or not
if (dp[n][k] == 1)
cout << "YES\n";
else
cout << "NO\n";
}
// Driver Code
int main()
{
int N = 2, M = 10, K = 5;
vector > arr = { { 4, 0, 15, 3, 2,
20, 10, 1, 5, 4 },
{ 4, 0, 10, 3, 2,
25, 4, 1, 5, 4 } };
PossibleSum(N, M, arr, K);
return 0;
} Java
// Java implementation to find
// whether is it possible to
// make sum equal to K
import java.io.*;
import java.util.*;
class GFG {
// Function that prints whether is it
// possible to make sum equal to K
static void PossibleSum(int n, int m,
int[][] v, int k)
{
int[][] dp = new int[n + 1][k + 1];
// Base case
dp[0][0] = 1;
for(int i = 0; i < n; i++)
{
for(int j = 0; j <= k; j++)
{
// Condition if we can make
// sum equal to current column
// by using above rows
if (dp[i][j] == 1)
{
// Iterate through current
// column and check whether
// we can make sum less than
// or equal to k
for(int d = 0; d < m; d++)
{
if ((j + v[i][d]) <= k)
{
dp[i + 1][j + v[i][d]] = 1;
}
}
}
}
}
// Printing whether is it
// possible or not
if (dp[n][k] == 1)
System.out.println("YES");
else
System.out.println("NO");
}
// Driver code
public static void main(String[] args)
{
int N = 2, M = 10, K = 5;
int[][] arr = new int[][]{ { 4, 0, 15, 3, 2,
20, 10, 1, 5, 4 },
{ 4, 0, 10, 3, 2,
25, 4, 1, 5, 4 } };
PossibleSum(N, M, arr, K);
}
}
// This code is contributed by coder001Python3
# Python3 implementation to find
# whether is it possible to
# make sum equal to K
# Function that prints whether is it
# possible to make sum equal to K
def PossibleSum(n, m, v, k):
dp = [[0] * (k + 1) for i in range(n + 1)]
# Base case
dp[0][0] = 1
for i in range(n):
for j in range(k + 1):
# Condition if we can make
# sum equal to current column
# by using above rows
if dp[i][j] == 1:
# Iterate through current
# column and check whether
# we can make sum less than
# or equal to k
for d in range(m):
if (j + v[i][d]) <= k:
dp[i + 1][j + v[i][d]] = 1
# Printing whether is it
# possible or not
if dp[n][k] == 1:
print("YES")
else:
print("NO")
# Driver Code
N = 2
M = 10
K = 5
arr = [ [ 4, 0, 15, 3, 2,
20, 10, 1, 5, 4 ],
[ 4, 0, 10, 3, 2,
25, 4, 1, 5, 4 ] ]
PossibleSum(N, M, arr, K)
# This code is contributed by divyamohan123C#
// C# implementation to find
// whether is it possible to
// make sum equal to K
using System;
class GFG{
// Function that prints whether is it
// possible to make sum equal to K
static void PossibleSum(int n, int m,
int[,] v, int k)
{
int[,] dp = new int[n + 1, k + 1];
// Base case
dp[0, 0] = 1;
for(int i = 0; i < n; i++)
{
for(int j = 0; j <= k; j++)
{
// Condition if we can make
// sum equal to current column
// by using above rows
if (dp[i, j] == 1)
{
// Iterate through current
// column and check whether
// we can make sum less than
// or equal to k
for(int d = 0; d < m; d++)
{
if ((j + v[i, d]) <= k)
{
dp[i + 1, j + v[i, d]] = 1;
}
}
}
}
}
// Printing whether is it
// possible or not
if (dp[n, k] == 1)
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
// Driver code
public static void Main(String[] args)
{
int N = 2, M = 10, K = 5;
int[,] arr = new int[,]{ { 4, 0, 15, 3, 2,
20, 10, 1, 5, 4 },
{ 4, 0, 10, 3, 2,
25, 4, 1, 5, 4 } };
PossibleSum(N, M, arr, K);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
YES时间复杂度： O(N * M * K)
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