[CareerCup] 9.8 Represent N Cents 组成N分钱 CareerCup Questions List 职业杯题目列表

9.8 Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), write code to calculate the number of ways of representing n cents.

给定一个钱数，用quarter，dime，nickle和penny来表示的方法总和。

Java:

public class Solution {
    /**
     * @param n an integer
     * @return an integer
     */
    public int waysNCents(int n) {
        int[] f = new int[n+1];
        f[0] = 1;
        int[] cents = new int[]{1, 5, 10, 25};
        for (int i = 0; i < 4; i++) 
            for (int j = 1; j <= n; j++) {
                if (j >= cents[i]) {
                    f[j] += f[j-cents[i]];
                }
            }
        return f[n];
    }
}　

Python:

class Solution:
    # @param {int} n an integer
    # @return {int} an integer
    def waysNCents(self, n):
        # Write your code here
        cents = [1, 5, 10, 25]
        ways = [0 for _ in xrange(n + 1)]
        
        ways[0] = 1
        for cent in cents:
            for j in xrange(cent, n + 1):
                ways[j] += ways[j - cent]
        
        return ways[n]

C++:

class Solution {
public:
    /**
     * @param n an integer
     * @return an integer
     */
    int waysNCents(int n) {
        // Write your code here
        vector<int> cents = {1, 5, 10, 25};
        vector<int> ways(n + 1);

        ways[0] = 1;
        for (int i = 0; i < 4; ++i)
            for (int j = cents[i]; j <= n; ++j)
                ways[j] += ways[j - cents[i]];

        return ways[n];
    }
};　　　　

C++:

class Solution {
public:
    int makeChange(int n) {
        vector<int> denoms = {25, 10, 5, 1};
        vector<vector<int> > m(n + 1, vector<int>(denoms.size()));
        return makeChange(n, denoms, 0, m);
    }
    int makeChange(int amount, vector<int> denoms, int idx, vector<vector<int> > &m) {
        if (m[amount][idx] > 0) return m[amount][idx];
        if (idx >= denoms.size() - 1) return 1;
        int val = denoms[idx], res = 0;
        for (int i = 0; i * val <= amount; ++i) {
            int rem = amount - i * val;
            res += makeChange(rem, denoms, idx + 1, m);
        }
        m[amount][idx] = res;
        return res;
    }
};

　　

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