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): 1846    Accepted Submission(s): 896 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,…

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…

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…

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…

分析 很多个较小的数字可以随机组合成较大的数字,所以B数组从小到大开始遍历,除了空集,最小的那个存在的个数对应的数字必然是a数组中的数字. 每求出这一部分之后,更新后续的B序列. 分析完后,主要的难点就是怎么去让已求出来的A序列随机组合,更新后续的B序列直接减就可以了.看成01背包问题,让m为背包去装 i,初始值为dp[0] = 1,由于i依次增大,A子集随机组合不会重复. dp[i]表示A序列中值为i的次数, 每次用第i个数的次数去更新i+1~m的dp值即可 #include <bits/st…

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 <…

快速数论变换ntt. 早上才刚刚接触了一下FFT,然后就开始撸这题了,所以要详细地记录一下. 看了这篇巨巨的博客才慢慢领会的:http://blog.csdn.net/cqu_hyx/article/details/52194696 FFT的作用是计算卷积.可以简单的理解为计算多项式*多项式最后得到的多项式,暴力计算是O(n*n)的,FFT可以做到O(nlogn). #pragma comment(linker, "/STACK:1024000000,1024000000") #inc…