=(x,y).因此满足dp的最优子结构条件 时间复杂度会有O(n*m) 详解和用树的思想分析符合dp 1func count( n, m )23forifrom0to n4forjfrom0to m5ifi equals06table[i, j] =17elseifj equals08table[i, j] =1ifi%S[j] equals0else09elseifS_j greater than i10table[i, j] = table[i, j ...
最终输出答案时做相应的简单求和即可。 代码 #include<iostream>#include<string.h>#include<stdio.h>#include<set>usingnamespacestd;#defineMAX_MONEY 250#defineCOIN_CNT 100longlongways[MAX_MONEY+5][COIN_CNT+5];intmain(intargc,constchar* argv[]){ set<int> coin={1,5,10,25,50};memset(ways,0...
two 5-cent coins and one 1-cent coin, one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent. ...
typedeflonglongll;constintmaxn=255;inta[5]={1,5,10,25,50}; ll dp[255][105];//dp[j][k]:用k个银币组成j值的个数intmain () {intx;while(cin>>x) { memset(dp,0,sizeof(dp)); dp[0][0]=1;for(inti=0;i<5;i++)for(intk=1;k<=100;k++)for(intj=a[i];j<=x;j++) dp...
UVA 674 Coin Change (DP) Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money. For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-...
Also, I would like to know generally how oneprovethat DP cannot solve some problem, if possible ? Thanks for reading You mean I will only need to add the condition x<=a[i] to while loop in this code which solve the problem when having inf. coins.Indeed. how stupid my question!
给出DP的递归求解公式: count[amount] = min{ count[amount - coins[i] ] } , 0 <= i < n; 这里count[amount] 表示组合出amount数值所需要的最小硬币数目,coins是硬币面值的数组。 根据这个公式,我们能轻松的给出伪代码: coinChange(int[] coins,intamount): ...
classSolution {public://无限背包问题intcoinChange(vector<int>& coins,intamount) { vector<int> dp(amount+1,amount+1); dp[0] =0;for(inti=1;i<= amount;i++){for(intj=0;j<coins.size();j++){//dp[i]初值都是amount+1if(i>=coins[j]) dp[i] = min(dp[i-coins[j]]+1,dp[i]...
Solutions of dynamic programming algorithm. Contribute to derailment/lintcode-dp development by creating an account on GitHub.
* You may assume that you have an infinite number of each kind of coin. * * @author zhangxu * @see https://leetcode.com/problems/coin-change/ * @see net.neoremind.mycode.argorithm.dp.CoinChange */ publicclassCoinChange{ } Copy lines Copy permalink...