两个角的硬币游戏（Greedy Approach）

考虑一个两个玩家的硬币游戏，其中每个玩家都一个一个地回合。有一排偶数硬币，轮到他/她的玩家可以从该排的两个角落中的任何一个角落中挑选一枚硬币。收集价值更高的硬币的玩家赢得游戏。为第一个回合的玩家制定策略，这样他/她就不会输掉比赛。

请注意，最多选择两个角的策略可能不起作用。在下面的例子中，第一个玩家在他/她使用策略最多选择两个角时输掉了比赛。

例子：

18 20 15 30 10 14
First Player picks 18, now row of coins is
  20 15 30 10 14
Second player picks 20, now row of coins is
  15 30 10 14
First Player picks 15, now row of coins is
  30 10 14
Second player picks 30, now row of coins is
  10 14
First Player picks 14, now row of coins is
  10 
Second player picks 10, game over.

The total value collected by second player is more (20 + 
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins. 

请注意，此问题与 Optimal Strategy for a Game | 不同。 DP-31。那里的目标是获得最大值。这里的目标是不松动。我们这里有一个贪心策略。这个想法是计算所有偶数硬币和奇数硬币的价值总和，比较这两个值。如果偶数硬币的总和更高，则首先移动的玩家始终可以确保其他玩家永远无法选择偶数硬币。类似地，如果奇数硬币的总和更高，他/她可以确保其他玩家永远无法选择奇数硬币。

例子：

18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64. 
Since the sum of even coins is more, the first 
player decides to collect all even coins. He first
picks 14, now the other player can only pick a coin 
(10 or 18). Whichever is picked the other player, 
the first player again gets an opportunity to pick 
an even coin and block all even coins. 

C++

// CPP program to find coins to be picked to make sure
// that we never loose.
#include 
using namespace std;
 
// Returns optimal value possible that a player can collect
// from an array of coins of size n. Note than n must be even
void printCoins(int arr[], int n)
{
    // Find sum of odd positioned coins
    int oddSum = 0;
    for (int i = 0; i < n; i += 2)
        oddSum += arr[i];
 
    // Find sum of even positioned coins
    int evenSum = 0;
    for (int i = 1; i < n; i += 2)
        evenSum += arr[i];
 
    // Print even or odd coins depending upon
    // which sum is greater.
    int start = ((oddSum > evenSum) ? 0 : 1);
    for (int i = start; i < n; i += 2)
        cout << arr[i] << " ";
}
 
// Driver program to test above function
int main()
{
    int arr1[] = { 8, 15, 3, 7 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    printCoins(arr1, n);
    cout << endl;
 
    int arr2[] = { 2, 2, 2, 2 };
    n = sizeof(arr2) / sizeof(arr2[0]);
    printCoins(arr2, n);
    cout << endl;
 
    int arr3[] = { 20, 30, 2, 2, 2, 10 };
    n = sizeof(arr3) / sizeof(arr3[0]);
    printCoins(arr3, n);
 
    return 0;
}

Java

// Java program to find coins to be
// picked to make sure that we never loose.
class GFG
{
 
// Returns optimal value possible
// that a player can collect from
// an array of coins of size n.
// Note than n must be even
static void printCoins(int arr[], int n)
{
// Find sum of odd positioned coins
int oddSum = 0;
for (int i = 0; i < n; i += 2)
    oddSum += arr[i];
 
// Find sum of even positioned coins
int evenSum = 0;
for (int i = 1; i < n; i += 2)
    evenSum += arr[i];
 
// Print even or odd coins depending
// upon which sum is greater.
int start = ((oddSum > evenSum) ? 0 : 1);
for (int i = start; i < n; i += 2)
    System.out.print(arr[i]+" ");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr1[] = { 8, 15, 3, 7 };
    int n = arr1.length;
    printCoins(arr1, n);
    System.out.println();
 
    int arr2[] = { 2, 2, 2, 2 };
    n = arr2.length;
    printCoins(arr2, n);
    System.out.println();
 
    int arr3[] = { 20, 30, 2, 2, 2, 10 };
    n = arr3.length;
    printCoins(arr3, n);
}
}
 
// This code is contributed by ChitraNayal

Python3

# Python3 program to find coins
# to be picked to make sure that
# we never loose
 
# Returns optimal value possible
# that a player can collect from
# an array of coins of size n.
# Note than n must be even
def printCoins(arr, n) :
 
    oddSum = 0
     
    # Find sum of odd positioned coins
    for i in range(0, n, 2) :
        oddSum += arr[i]
 
    evenSum = 0
     
    # Find sum of even
    # positioned coins
    for i in range(1, n, 2) :
        evenSum += arr[i]
 
    # Print even or odd
    # coins depending upon
    # which sum is greater.
    if oddSum > evenSum :
        start = 0
    else :
        start = 1
 
    for i in range(start, n, 2) :
        print(arr[i], end = " ")
 
# Driver code
if __name__ == "__main__" :
     
    arr1 = [8, 15, 3, 7]
    n = len(arr1)
    printCoins(arr1, n)
    print()
     
    arr2 = [2, 2, 2, 2]
    n = len(arr2)
    printCoins(arr2, n)
    print()
     
    arr3 = [20, 30, 2, 2, 2, 10]
    n = len(arr3)
    printCoins(arr3, n)
     
# This code is contributed by ANKITRAI1

C#

// C# program to find coins to be
// picked to make sure that we never loose.
using System;
 
class GFG
{
 
// Returns optimal value possible
// that a player can collect from
// an array of coins of size n.
// Note than n must be even
static void printCoins(int[] arr, int n)
{
     
// Find sum of odd positioned coins
int oddSum = 0;
for (int i = 0; i < n; i += 2)
    oddSum += arr[i];
 
// Find sum of even positioned coins
int evenSum = 0;
for (int i = 1; i < n; i += 2)
    evenSum += arr[i];
 
// Print even or odd coins depending
// upon which sum is greater.
int start = ((oddSum > evenSum) ? 0 : 1);
for (int i = start; i < n; i += 2)
    Console.Write(arr[i]+" ");
}
 
// Driver Code
public static void Main()
{
    int[] arr1 = { 8, 15, 3, 7 };
    int n = arr1.Length;
    printCoins(arr1, n);
    Console.Write("\n");
 
    int[] arr2 = { 2, 2, 2, 2 };
    n = arr2.Length;
    printCoins(arr2, n);
    Console.Write("\n");
 
    int[] arr3 = { 20, 30, 2, 2, 2, 10 };
    n = arr3.Length;
    printCoins(arr3, n);
}
}
 
// This code is contributed by ChitraNayal

PHP

 $evenSum) ? 0 : 1);
    for ($i = $start; $i < $n; $i += 2)
        echo $arr[$i]." ";
}
 
// Driver Code
$arr1 = array( 8, 15, 3, 7 );
$n = sizeof($arr1);
printCoins($arr1, $n);
echo "\n";
 
$arr2 = array( 2, 2, 2, 2 );
$n = sizeof($arr2);
printCoins($arr2, $n);
echo "\n";
 
$arr3 = array( 20, 30, 2, 2, 2, 10 );
$n = sizeof($arr3);
printCoins($arr3, $n);
 
// This code is contributed by ChitraNayal
?>

Javascript

 
// Javascript program to find coins to
// be picked to make sure that we never
// loose.
 
// Returns optimal value possible that
// a player can collect from an array
// of coins of size n. Note than n must be even
function printCoins(arr, n)
{
     
    // Find sum of odd positioned coins
    var oddSum = 0;
    for(var i = 0; i < n; i += 2)
        oddSum += arr[i];
 
    // Find sum of even positioned coins
    var evenSum = 0;
    for(var i = 1; i < n; i += 2)
        evenSum += arr[i];
 
    // Print even or odd coins depending upon
    // which sum is greater.
    var start = ((oddSum > evenSum) ? 0 : 1);
    for(var i = start; i < n; i += 2)
        document.write(arr[i] + " ");
}
 
// Driver code
var arr1 = [ 8, 15, 3, 7 ]
var n = arr1.length;
printCoins(arr1, n);
document.write("<br>");
 
var arr2 = [ 2, 2, 2, 2 ]
var n = arr2.length;
printCoins(arr2, n);
document.write("<br>");
 
var arr3 = [ 20, 30, 2, 2, 2, 10 ]
n = arr3.length;
printCoins(arr3, n);
 
// This code is contributed by noob2000
 

输出：

15 7 
2 2 
30 2 10

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