DP7 两道换硬币的问题 Coin Change @geeksforgeeks

1 Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter.

两种思路：

1）http://www.cs.ucf.edu/~dmarino/ucf/cop3530/lectures/COP3530DynProg02.doc

特别注意到换完的硬币是没有顺序的，所以要impose order

2）http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/

http://www.cnblogs.com/python27/p/3303721.html

http://www.mathblog.dk/project-euler-31-combinations-english-currency-denominations/

考虑是否取最后一个硬币，转换为背包问题

package DP;import java.util.Arrays;/**Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter. */public class CoinChange1 {public static int MEM[] = new int[10001];   // Can support up to 10000 peso value         public static void main(String[] args) {        int coins[] = {1, 2, 3};  // Available coin denominations    Arrays.sort(coins);    int coinKinds = coins.length;    int sum = 4;    System.out.println(countRec_1st(coins, coinKinds, sum, coins[coins.length-1]));    System.out.println(countRec_2nd(coins, coinKinds, sum));    System.out.println(countDP_1st(coins, coinKinds, sum));    System.out.println(countDP2D_2nd(coins, coinKinds, sum));    System.out.println(countDP1D_2nd(coins, coinKinds, sum));    }      //====================================第一种思路    /*     max 用来保持order，否则就会有重复，如(1,2),(2,1)     */    public static int countRec_1st(int coins[], int coinKinds, int sum, int max){    if(sum < 0){    return 0;    }    if(sum == 0){    return 1;    }        int ways = 0;    for(int i=0; i<coinKinds; i++){    if(max >= coins[i]){    ways += countRec_1st(coins, coinKinds, sum-coins[i], coins[i]);    }    }    return ways;    }            // http://www.cnblogs.com/python27/p/3303721.html    // dp[i][j] = sum(dp[i-1][j-k*coins[i-1]]) for k = 1,2,..., j/coins[i-1]    // dp[0][j] = 1 for j = 0, 1, 2, ..., sum    public static int countDP_1st(int[] coins, int coinKinds, int sum){    int[][] dp = new int[coinKinds+1][sum+1];        for(int i=0; i<=coinKinds; i++){    for(int j=0; j<=sum; j++){    dp[i][j] = 0;    }    }    for(int i=0; i<=coinKinds; i++){    dp[i][0] = 1;    }        for(int i=1; i<=coinKinds; i++){// 币的面值    for(int j=1; j<=sum; j++){// 要凑成的数量    dp[i][j] = 0;    for(int k=0; k<=j/coins[i-1]; k++){    dp[i][j] += dp[i-1][j-k*coins[i-1]];    }    }    }        return dp[coinKinds][sum];    }    //====================================第二种思路        // Return the count of ways we can sum coins[0...m-1] coins to get sum     public static int countRec_2nd(int coins[], int coinKinds, int sum){    if(sum == 0){// If n is less than 0 then no solution exists    return 1;    }    if(sum < 0){// If n is less than 0 then no solution exists    return 0;    }    if(coinKinds<=0 && sum>=1){// If there are no coins and n is greater than 0, then no solution exist    return 0;    }    // count is sum of solutions (i) including coins[m-1] (ii) excluding coins[m-1]    // 两种情况：    // 1. 不使用最后一个硬币，因此实际上就是用少了最后一个硬币的coins来凑sum    // 2.使用最后一个硬币，因此sum的总数减少，因为每个硬币都有无数个，所以m不变    return countRec_2nd(coins, coinKinds-1, sum) + countRec_2nd(coins, coinKinds, sum-coins[coinKinds-1]);    }            public static int countDP2D_2nd(int[] coins, int coinKinds, int sum){    // We need n+1 rows as the table is constructed in bottom up manner using         // the base case 0 value case (n = 0)    int[][] dp = new int[sum+1][coinKinds];        // Fill the entries for 0 value case (sum = 0)    for(int i=0; i<coinKinds; i++){    dp[0][i] = 1;    }        // Fill rest of the table entries in bottom up manner      for(int i=1; i<=sum; i++){// sum    for(int j=0; j<coinKinds; j++){// coins m    // Count of solutions including S[j]    int x = (i-coins[j]>=0) ? dp[i-coins[j]][j] : 0;// 要满足i-coins[j]>=0    // Count of solutions excluding S[j]    int y = (j>=1) ? dp[i][j-1] : 0;// 要满足j-1>=0    dp[i][j] = x + y;    }    }    return dp[sum][coinKinds-1];    }            public static int countDP1D_2nd(int[] coins, int coinKinds, int sum){        // dp[i] will be storing the number of solutions for        // value i. We need n+1 rows as the table is constructed        // in bottom up manner using the base case (n = 0)    int[] dp = new int[sum+1];        dp[0] = 1;// Base case (If given value is 0)    // Pick all coins one by one and update the table[] values        // after the index greater than or equal to the value of the        // picked coin    for(int i=0; i<coinKinds; i++){// coins m    for(int j=coins[i]; j<=sum; j++){// sum    dp[j] += dp[j-coins[i]];    }    }    return dp[sum];    }    }

2 What is the smallest number of coins to change? 如果换，最少会有多少枚硬币？

http://www.columbia.edu/~cs2035/courses/csor4231.F11/dynamic.pdf

C[p] = min{C[p-di]+1} 

package DP;import java.util.Arrays;/**A country has coins with denominations1 = d1 < d2 < · · · < dk.  You want to make change for n cents, using the smallest number  */public class CoinChange2 {public static int MEM[] = new int[10001];   // Can support up to 10000 peso value    public static int coins[] = {1, 2, 3};  // Available coin denominations         public static void main(String[] args) {    int n = 4;        System.out.println(minCoins(n));        System.out.println(minCoins_DP(n));    }            // 记忆化搜索，top-down 递归    public static int minCoins(int n) {        if(n < 0)            return Integer.MAX_VALUE -1;        else if(n == 0)            return 0;        else if(MEM[n] != 0)    // If solved previously already            return MEM[n];        else {            // Look for the minimal among the different denominations            MEM[n] = 1+minCoins(n-coins[0]);            for(int i = 1; i < coins.length; i++)                MEM[n] = Math.min(MEM[n], 1+minCoins(n-coins[i]));            return MEM[n];        }    }    // bottom-up DPpublic static int minCoins_DP(int n){int[] minCoins = new int[n+1];Arrays.fill(minCoins, Integer.MAX_VALUE);// 第一个硬币minCoins[0] = 0;// 算出n前的每一个可换硬币的数量for(int i=1; i<=n; i++){// 根据递推公式，看看硬币可拆分的可能性for(int j=0; j<coins.length; j++){if(coins[j] <= i){minCoins[i] = Math.min(minCoins[i], 1+minCoins[i-coins[j]]);}}}return minCoins[n];}}