Given a set of numbers [1-N] . Find the number of subsets such that the sum of numbers in the subset is a prime number.
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Maximum sum can be = N*(N+1) / 2 for
any subset considered.
Hence put S = (N*(N+1) / 2) in subset sum problem.
i.e dp[i][j] = dp[i-1][j] + dp[i-1][j-a[i]]; // check for cases when j-a[i] < 0.
iterate i from 1 to N and j from 0 to S.
ans = 0;
again iterate from j = 0 to S and check if j is prime, and if j is prime then
ans = ans + dp[N][j];
output ans.
Dynamic Programming is often associated with the number of blank blank....

本文探讨了如何利用动态规划解决一个有趣的问题:给定一系列从1到N的数字,找出所有子集中元素之和为质数的子集数量。通过将最大可能的子集和作为输入,使用动态规划进行计算,并最终统计得到所有符合条件的子集数目。
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