然后开一个大小为k的数组ans,初始化为1.并且将n减去k。这时候只需要k个数字不断往上翻倍即可。 遍历数组ans,遍历每个元素。for(i,0:k)while(只要当前第i个元素ans[i]小于n,则使得n减去ans[i],然后当前元素a*=2) 如果最后n还是不为0,那么就输出no,否则把数组ans输出即可。 我就不把我当时写的可长的...
CodeForces - 702B Powers of Two B. Powers of Two time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output You are given n integers a1, a2, ..., an...Codeforces Round #529 -C- Powers Of Two(二进制拆分) A positive integer xx is...
codeforces Powers of Two 小白错误总结之暴int: 题目大意:给你一系列的数字,拿出两个进行匹配,看是否是2的幂,输出总匹配数;数据范围1e9,长度1e5; 想法:匹配分为跟自己匹配和跟另一个数进行匹配,自己配就是排列n找2,n!/(n-2)!,不同直接乘。 考虑到数字相同的情况,就要考虑到压缩,排序(sort,o(n*log(...
Points and Powers of Two CodeForces - 988D There are n distinct points on a coordinate line, the coordinate of i-th point equals to xi. Choose a subset of the given set of points such that the distance between each pair of points in a subset i......
CodeForces 702B Powers of Two (暴力,优化) 代码人生 题意:给定 n 个数,问你从有多少下标 i < j,并且 ai + aj 是2的倍数。 析:方法一: 从输入开始暴力,因为 i < j 和 i > j 是一样,所以可以从前面就开始查找,然后计数,用个map就搞定,不过时间有点长,接近两秒。
codeforces-Powers of Two(map的应用) 应用#includei++其他文章分类代码人生 B. Powers of Two time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are givennintegersa1, a2, ..., an. Find the number of pairs of indexesi, ...
Ivan has got an array of n non-negative integers a1, a2, ..., an. Ivan knows that the array is sorted in the non-decreasing order. Ivan wrote out integers 2a1, 2a2, ..., 2an on a piece of paper. Now he wonders, what minimum number of integers of form 2b (b...
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