Given any 10 different positive integers less than 100.there will be two disjoint subsets with the same sum.举个例子我任意给出10个:44,54,84,24,14,95,67,59,35,48。这10个数的集合里会有2个不相交的子集,他
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/27 14:24:40
Given any 10 different positive integers less than 100.there will be two disjoint subsets with the same sum.举个例子我任意给出10个:44,54,84,24,14,95,67,59,35,48。这10个数的集合里会有2个不相交的子集,他
Given any 10 different positive integers less than 100.there will be two disjoint subsets with the same sum.
举个例子我任意给出10个:44,54,84,24,14,95,67,59,35,48。这10个数的集合里会有2个不相交的子集,他们的和相同比如44,24和54,14。这个是证明题
Given any 10 different positive integers less than 100.there will be two disjoint subsets with the same sum.举个例子我任意给出10个:44,54,84,24,14,95,67,59,35,48。这10个数的集合里会有2个不相交的子集,他
给定任意10个小于100的正整数.会有两个不相交的子集有相同的和.(感觉逻辑不大通.这两句话本来就是连在一道的么?还是当中有省略的部分呀)如32,42,62,22,12,67,56,33,41
32,42,62,22,12,67,56,33,41。
给定任意10个小于100的正整数。会有两个不相交的子集有相同的和。(感觉逻辑不大通。这两句话本来就是连在一道的么?还是当中有省略的部分呀)
打电话问老师吧(老师很喜欢爱提问的学生哟)O(∩_∩)O哈哈~