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 integerdef waysNCents(self, n):# Write your code herecents = [1, 5, 10, 25]ways = [0 for _ in xrange(n + 1)]ways[0] = 1for 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 herevector<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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转载于:https://www.cnblogs.com/lightwindy/p/8674187.html

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