Time limit1000 ms Memory limit65536 kB The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the ...
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 16213 Accepted Submission(s): 4992 Problem Descr ...
题目描述身上有p块钱,可以连续购买k件物品只需要付其中最贵的价钱,求出能够购买物品的最大数量思路我们用f[i]f[i]f[i]表示购买前i件商品的最小花费,因为可以连续购买k件物品,所以当i大于k时,得到转移方程f[i]=min(f[i−1],f[i−k])+a[i]f[i]=min(f[i-1],f[i-k])+a[i]f[i]=min(f[i−1],f[i−k])+a[i]。AC代码#include#define x first#define y