题目:http://poj.org/problem?id=1742 贪心地想,1.如果一种面值已经可以被组成,则不再对它更新: 2.对于同一种面值的硬币,尽量用较少硬币(一个)更新,使后面可以用更多此面值硬币. 代码如下: #include<iostream> #include<cstdio> #include<cstring> using namespace std; ],c[],m,used[],ans; ]; int main() { ) { ; memset(f…

Coins Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 34814   Accepted: 11828 Description 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…

Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 32971   Accepted: 11950 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The m…

Vjudge传送门 $Sol$ 首先发现这是一个多重背包,所以可以用多重背包的一般解法(直接拆分法,二进制拆分法...) 但事实是会TLE,只能另寻出路 本题仅关注“可行性”(面值能否拼成)而不是“最优性”,这是一个特殊之处. 从这里找优化 在“最优性”的问题中,$f[j]$从$f[j]$或$f[j-a[i]]$中转移而来:而在这样的“可行性”问题中,其实只要$f[j]$可行,我们就可以不用考虑$f[j-a[i]$了,也可以反过来说. 于是我们可以考虑一种贪心策略,设$used[j]$表示$f[…

意甲冠军:你有N种硬币,每个价格值A[i],每个号码C[i],要求. 在不超过M如果是,我们用这些硬币,有多少种付款的情况下,.那是,:1,2,3,4,5,....,M这么多的情况下,,你可以用你的硬币不找零,种情况. 比如: 你有一种硬币,价值2.个数2,那么 你是不能付款 3元的..你仅仅能付款2,或者4元.. OK,题意差点儿相同就是这样啦. 那么这里有两种方式! 分析: 那么这里我们能够用多重背包来解决,我们把价值和重量看成一样的w[i] = A[i]:用M作为背包. 那么dp 过后.我…

Dividing Time Limit: 1000MS Memory Limit: 10000K Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had…

Description 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 pri…

Dividing Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 68769   Accepted: 17955 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbl…

Description 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 pri…

题意:有n种面额的硬币.面额.个数分别为A_i.C_i,求最多能搭配出几种不超过m的金额? 思路:dp[j]就是总数为j的价值是否已经有了这种方法,如果现在没有,那么我们就一个个硬币去尝试直到有,这种价值方法有了的话,那么就是总方法数加1.多重背包可行性问题 传统多重背包三重循环会超时,因为只考虑是否可行,没有考虑剩余面额数量的因素. o(n*v)方法 #include <iostream> #include <cstdio> #include <string.h> #…