UVA 10317 - Equating Equations (背包)】的更多相关文章

Problem F Equating Equations Input: standard input Output: standard output Time Limit: 6 seconds Memory Limit: 32 MB Read equations with up to 16 terms and + and – operators (not unary) and reorder the terms (but not the operators) so that the equati…
UVA 10306 e-Coins(全然背包: 二维限制条件) option=com_onlinejudge&Itemid=8&page=show_problem&problem=1247">http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1247 题意: 对于每一个例子.先给定两个数n,m,分别表示有…
题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=565 题意很好理解,普通的01背包,dp[i - 1][j]表示在前i - 1件物品中选取若干物品放入容量为j背包所得到的最大的价值,dp[i - 1][j - w[i]] + v[i]表示前i - 1件物品中选取若干物品放入容量为j - w[i]背包所得到的最大的价值加上第i件物…
People in Cubeland use cubic coins. Not only the unit of currency iscalled a cube but also the coins are shaped like cubes and their valuesare cubes. Coins with values of all cubic numbers up to 9261(= 213),i.e., coins with the denominations of 1, 8,…
题目连接:147 - Dollars 题目大意:有11种硬币, 现在输入一个金额, 输出有多少种组成方案. 解题思路:uva 674 的升级版,思路完全一样, 只要处理一下数值就可以了. #include <stdio.h> #include <string.h> const int N = 30005; const int val[11] = {5, 10, 20, 50, 100, 200, 500, 1000, 2000, 5000, 10000}; long long n,…
//路径记录方法:若是dp[j-value[i]]+value[i]>dp[j]说明拿了这个东西,标志为1, //for循环标志,发现是1,就打印出来,并把背包的容量减少,再在次容量中寻找标志: #include <iostream> #include <cstring> #include <algorithm> using namespace std; ],dp[],s[][]; int main() { int n,m; while(cin>>m)…
DescriptionCD You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most…
这里就是01背包多了一维物品个数罢了 记得不能重复所以有一层循环顺序要倒着来 边界f[0][0] = 1 #include<cstdio> #include<vector> #include<cstring> #define REP(i, a, b) for(int i = (a); i < (b); i++) using namespace std; const int MAXN = 1121; bool is_prime[MAXN]; vector<in…
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<ctype.h> #include<algorithm> #define N 1010 using namespace std; int dp[N], path[N][N], w[N]; int main() { int v, n; while(~scanf("%d", &v)) { sca…
// 集训最终開始了.来到水题先 #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; int a[23]; int d[23][100000]; int flag[23]; int W,n; void init(){ cin >> n; for (int i=1;i<=n;i++) cin >…