题目链接：POJ 1742 Coins 题目大意： 给出硬币面额及每种硬币的个数，求从\(1\)到\(m\)能凑出面额的个数。 题解： 设\(dp[i][j]\)为前\(i\)种面值的硬币组成面额\(j\)时，第\(i\)种硬币剩余的数量。 默认\(dp[i][j] = -1\)，表示无法组成面额\(j\)。 若\(dp[i - 1][j]\geq 0\)，则说明已经可以
题目链接：http://poj.org/problem?id=1742 与一般的背包问题不一样，这是要计算满足条件的情况的数量，而不是计算最值，一开始的思路就是按照书上的类比： dp[i][j] := 用前i种硬币能否凑成j 递推：dp[i][j] = （dp[i – 1][j – k * A[i]]）为真的时候 但是 MLE，其实一点都不惊讶吧，数组开那么
People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change
People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change
给出硬币面额及每种硬币的个数，求从1到m能凑出面额的个数。 简单的多重部分和问题，动态规划解决 递推式可以参考https://mp.csdn.net/postedit/98054811 #include<iostream> #include<cstdio> #include<fstream> #include<queue> #include<vector> #include<algorithm> #include
