题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1085 解题报告:有1,2,5三种面值的硬币,这三种硬币的数量分别是num_1,num_2,num_5,问你不能凑的钱的最小值是多少. DP,开一个这样的数组dp[i][3],然后dp[i][0]表示凑成 i 元钱需要dp[i][0]张1块的,需要dp[i][1]张两块的,dp[i][2]张5块的,然后依次往后递推,得到 i 的途径一共有三种,第一种是i-1加一张一块的,第二种是i-2加上1张两块的,…

Holding Bin-Laden Captive! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17653    Accepted Submission(s): 7902 Problem Description We all know that Bin-Laden is a notorious terrorist, and he h…

Holding Bin-Laden Captive! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13861    Accepted Submission(s): 6230 Problem Description We all know that Bin-Laden is a notorious terrorist, and he…

Problem Description We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! "Oh, God! How terrible! " Don't be so afraid, guys. Although he hi…

Problem Description We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! “Oh, God! How terrible! ” Don’t be so afraid, guys. Although he hides i…

题意: 有面值分别为1.2.5的硬币,分别有num_1.num_2.num_5个,问不能组成的最小面值是多少?(0<=每种硬币个数<=1000,组成的面值>0) 思路: 母函数解决.只有3个括号要作乘法,分别代表面值1.2.5所能组成的情况.需要两个数组,所能组成的最大值为num_1+2*num_2+5*num_5.如果在这个范围内都能组成,那么最小不能组成的面值为num_1+2*num_2+5*num_5+1.若没有1分钱的硬币,那么不能组成的肯定是1了. 数组的用法:ans[]保存第…

生成函数题. 题意:有币值1,2,5的硬币若干,问你最小的不能组成的币值为多少. 解法:写出生成函数: 然后求每项的系数即可. 因为三种硬币最多1000枚,1*1000+2*1000+5*1000=8000,那么多项式乘积的最高次数为8000 用c保存累计相乘各项的系数,tc保存c和当前项相乘的系数 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> us…

//给你面值为1,2,5的三种硬币固定的数目,求不能凑出的最小钱数 //G(x)=(1+x+...+x^num1)(1+x^2+...+x^2num2)(1+x^5+,,,+x^5num3), //展开,系数不为0的数都是能够由硬币组合出来的. # include <algorithm> # include <string.h> # include <stdio.h> # include <iostream> using namespace std; int…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1693 第一道插头 DP ! 直接用二进制数表示状态即可. #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; ,M=(<<)+; int n,m,bin[N];ll dp[N][N][M]; int b[N][N]; void…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…