POJ-2229 Sumsets---完全背包变形

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poj 2229 Sumsets 完全背包求方案总数

Sumsets Description Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 1) 1+1+1+1+1+1+1 2)…

poj 2229 【完全背包dp】【递推dp】

poj 2229 Sumsets Time Limit: 2000MS Memory Limit: 200000K Total Submissions: 21281 Accepted: 8281 Description Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an…

POJ 2229 Sumsets(技巧题, 背包变形)

discuss 看到有人讲完全背包可以过, 假如我自己做的话, 也只能想到完全背包了思路: 1. 当 n 为奇数时, f[n] = f[n-1], 因为只需在所有的序列前添加一个 1 即可, 所有的序列同时延迟 1 位, 不会出现重复若是这个 1 和其他的1组成 2 而不是放在首位, 怎么办? 不会这样, 因为这个序列肯定已经存在了证明, 假设sum(s1) = 2*k, s1内部某个1加1得到 s2, 则 sum(s2) = 2*k+1, s2 的首位仍然肯定是1, 那么 s2 也可以通…

poj -2229 Sumsets （dp）

http://poj.org/problem?id=2229 题意很简单就是给你一个数n,然后选2的整数幂之和去组成这个数.问你不同方案数之和是多少? n很大,所以输出后9位即可. dp[i] 表示组成i的不同方案数,那么 dp[1]=1;dp[2]=2; if(i%2) dp[i]=dp[i-1] ; i如果是奇数,那么只能在i-1的每个方案数前面加上1得到i,所以方案数相等. else dp[i]=dp[i-1]+dp[i/2] ; i如果是偶数,一种可能是i有两个1,在i-1的每个方案…

POJ 2229 Sumsets

Sumsets Time Limit: 2000MS Memory Limit: 200000K Total Submissions: 11892 Accepted: 4782 Description Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer…

poj 2229 Sumsets（dp）

Sumsets Time Limit : 4000/2000ms (Java/Other) Memory Limit : 400000/200000K (Java/Other) Total Submission(s) : 4 Accepted Submission(s) : 3 Problem Description Farmer John commanded his cows to search for different sets of numbers that sum to a g…

POJ 2923 Relocation(01背包变形, 状态压缩DP)

Q: 如何判断几件物品能否被 2 辆车一次拉走? A: DP 问题. 先 dp 求解第一辆车能够装下的最大的重量, 然后计算剩下的重量之和是否小于第二辆车的 capacity, 若小于, 这 OK. Description Emma and Eric are moving to their new house they bought after returning from their honeymoon. Fortunately, they have a few friends helping…