Rikka with Subset Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1846    Accepted Submission(s): 896 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation…

dp[i][j]表示前i个元素,子集和为j的个数.d[i][j] = d[i][j] + d[i-1][j-k] (第i个元素的值为k).这里可以优化成一维数组 比如序列为 1 2 3,每一步的dp值为 1 0 0 0 0 0 0 (d[0][0]=1) 1 1 0 0 0 0 0 1 1 1 1 0 0 0 1 1 1 2 1 1 1 最终的序列就是题目给出的B序列,把B序列减去每一次dp得到的序列,第一个非0值就是a序列中的值 #include <cstdio> #include <…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739    Accepted Submission(s): 2470 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…

Rikka with Subset Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1122    Accepted Submission(s): 541 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation…

Rikka with Subset Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 837    Accepted Submission(s): 411 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation,…

题意:求解01背包价值的第K优解. 分析: 基本思想是将每个状态都表示成有序队列,将状态转移方程中的max/min转化成有序队列的合并. 首先看01背包求最优解的状态转移方程:\[dp\left[ j \right] = \max \left\{ {dp\left[ j \right],dp\left[ {j - a\left[ i \right].w} \right] + a\left[ i \right].v} \right\}\] 如果要求第K优解,那么状态 dp[j] 就应该是一个大小为…

Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has n positive A1−An and their sum is m. Then for each subset S of A, Yuta calcula…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6092 #include <cstdio> #include <iostream> #include <cstring> using namespace std; ; int b[maxn] , dp[maxn] , n , m; int main() { int t; scanf("%d", &t); while(t--) { scanf(&qu…

http://acm.hdu.edu.cn/showproblem.php?pid=6092 题意: 给出两个数组A和B,A数组一共可以有(1<<n)种不同的集合组合,B中则记录了每个数出现的次数,现在要根据B数组来推出A数组最小的序列. 思路: 如果$B_{i}$是 B 数组中除了$B_{0}$ 以外第一个值不为 0 的位置,那么显然 i 就是 A 中的最小数. 那么我们每次取出$B_{i}$一个数,对于后面的数组来说,满足$B_{j}=B_{j}-B_{j-i}$,为什么? 其实仔细想想就…

题目链接 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has n positive A1−An and their sum is m. Then for each subset S of A, Yuta ca…