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POJ 1742 Coins
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POJ 1742(Coins)
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Coins POJ - 1742
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 changeCoins POJ - 1742
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 changeCoins POJ - 1742
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