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The past couple of months have seen an increased acknowledgement of the role social media has to play in the development of the financial services sector.  Industry forums like Finextra's Live Social Media Days held in London and New York and the ina…
Investment Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu   Description John never knew he had a grand-uncle, until he received the notary��s letter. He learned that his late grand-uncle had gathered a lot of money, somew…
Investment Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8019   Accepted: 2747 Description John never knew he had a grand-uncle, until he received the notary's letter. He learned that his late grand-uncle had gathered a lot of money, s…
A - Investment Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2063 Description John never knew he had a grand-uncle, until he received the notary's letter. He learned that his late grand-uncle…
A - Investment John never knew he had a grand-uncle, until he received the notary's letter. He learned that his late grand-uncle had gathered a lot of money, somewhere in South-America, and that John was the only inheritor. John did not need that muc…
题目链接:http://poj.org/problem?id=2063 今天果然是卡题的一天.白天被hdu那道01背包的变形卡到现在还没想通就不说了,然后晚上又被这道有个不大也不小的坑的完全背包卡了好久.这题主要是说让你选择不同的债券(每种债券的费用和年利率的属性),然后n年后使得本金最大,如果仅仅是问一年的话就是个裸完全背包的题了,不过它是n年,每年得到的总利息都会加入到本金中变为下一年的本金,知道了这个后就很好处理了,在这道题里每年变化的本金就是背包容量,然后债券那两个属性就是物品的费用(o…
题目链接:http://poj.org/problem?id=2063 题意:银行每年提供d种债券,每种债券需要付出p[i]块钱,然后一年的收入是v[i],到期后我们把本金+收入取出来作为下一年度本金继续购买债券. 问你起始本金为n元,m年后你手里最多能有多少钱. 其实这题不难..我却想了一会.. 因为题目保证了p[i]是1000的倍数,所以我们可以把本金和p[i]都先除以1000,然后算出每年可能获得的最大收入,然后加到本金当中,在暴力枚举m年就行了. 设计状态dp[j]代表我花了不超过j元钱…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1963 //多重背包 #include <cstdio> #include <cstring> #include <iostream> using namespace std; + ; #define N 15 long long dp[maxn], ans; int c[N], w[N], V; void Pack(int C, int W) { for(int i = C…
题意:给定一个初始资金capital,然后给定d种投资方案,每种投资方案中有投资额value[i](是1000的倍数)和利息interest[i],每年的投资就可以拿到全部利息,然后累加起来继续投资利滚利.问经过year年后最多拥有多少资金?// 一年年的求最大利润就可以了// 每一年内都是完全背包// 这题的关键是value[i]都是1000的倍数 就可以把空间除以1000 了不然时间复杂度就高了好多#include <iostream> #include <algorithm>…
完全背包. #include <stdio.h> #include <string.h> #define max(a, b) (a>b) ? a:b ], an[]; ]; int main() { int case_n; int n, m, year; int i, j, k; scanf("%d", &case_n); while (case_n--) { scanf("%d %d", &m, &year);…