完全背包模板，和01背包相比不用倒推，因为一种可以选多个。 这道题求最小，dp数组初始化为无穷即可。 1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int M=10005; 6 const int INF=0x3f3f3f; 7 int dp[M];//dp[j]表示放入
#include<bits/stdc++.h> using namespace std; int dp[10005]; int data77[505][2]; int minhh(int x,int y) { if(x<=y)return x; else return y; } int main() { int n,E,F,N,i,j; scanf("%d",&n); while(n--) {
Piggy-Bank Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1114 Problem Description Before ACM can do anything, a budget must be prepared and the necessary financial support
问题：在编译Android 系统kernel源码的时候出现错误 I am trying to build android project Getting below error even after setting correct path for toolchains /bin/sh: lzop: not found AS arch/arm/boot/compressed/piggy.lzo.o arch/arm/boot/compressed/piggy
http://poj.org/problem?id=1384 完全背包是物品的个数是无限的。 二维数组：（这个题目用二维数组会超时） 由于dp数组初始化的问题，一直答案不对。 后面发现是： for(int i=0;i<=N;i++) dp[i][0]=0; #include<cstdio> #include<iostream> #include<string> #include<cstring
题意 首先给出\(e\)和\(f\)，代表空猪重量和空猪装满硬币的重量； 接下去给出1个\(n\)，表示接下来又\(n\)行不同的硬币类型，每种硬币再给出一个价值\(p\)和重量\(w\)； 最后问我们是否能输出存钱罐猪中的最小金额，如果可以直接输出给定格式的语句，否则输出 This is impossible. 。 思路 完全
HDU - 1114 题目翻译： 在acm能够做任何事情之前, 必须编制预算并获得必要的财政支持。这一行动的主要收入来自IBM。这个想法其实很简单,每当一些会员有一点小钱时,他就会把所有的硬币都扔到小猪存钱罐里。这个过程是不可逆转的, 除非打破猪，否则硬币不能拿出来。过了足够长的时间,
Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he take
#include<iostream> #include<cstring> using namespace std; const int INF=0x3f3f3f3f; int t,e,f,n,dp[10010]; struct node { int p, w; //价值 重量 } no[10000]; int main() { cin >> t; while(t--) { memset(dp,0x3f,sizeof dp);
Description Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small
Piggy-Bank Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 37378 Accepted Submission(s): 18528 Problem Description Before ACM can do anything, a budget must be prepared and the necessary finan
