CF439C Devu and Partitioning of the Array 题解 以下用evev表示奇数堆的个数,odod表示偶数堆的个数,不考虑无解的情况。 首先这里有三个式子: 奇数+奇数=偶数奇数+奇数=偶数 偶数+偶数=偶数偶数+偶数=偶数 奇数+偶数=奇数奇数+偶数=奇数 所以我们可以看见,对于奇数堆而言,我们需要有奇数个奇数来填充每一堆,而这是 CF 的构造题,于
1.让两个偶数集合合并在一起,变成一个更大的偶数集合,v[0].size()-- 2.让一个偶数集合和一个奇数集合合并在一起,变成一个奇数集合,v[0].size()--,但是奇数集合个数还是不变 此后v[0].size==p了 那么接下来只要调整v[1].size就好了 但是在调整之前先判断v[0].size()+v[1].size()>=...
If it is possible to partition the array, also give any possible way of valid partitioning. Input The first line will contain three space separated integersn,k,p(1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k). The next line will containnspace-separated distinct integers repr...
1.让两个偶数集合合并在一起,变成一个更大的偶数集合,v[0].size()-- 2.让一个偶数集合和一个奇数集合合并在一起,变成一个奇数集合,v[0].size()--,但是奇数集合个数还是不变 此后v[0].size==p了 那么接下来只要调整v[1].size就好了 但是在调整之前先判断v[0].size()+v[1].size()>=...
Devu and Partitioning of the Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which ...
Devu and Partitioning of the Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which ...
题目链接:Codeforces 439C Devu and Partitioning of the Array 题目大意:给出n个数,要分成k份,每份有若干个数,可是仅仅须要关注该份的和为奇数还是偶数,要求偶数堆的个数为p。 输出方案。 解题思路:首先先将数组依照奇偶排序。也能够分开储存。 然后先单独分k-p个奇数,然后后面的就将两个奇数当一个偶数分配。
C. Devu and Partitioning of the Array time limit per test memory limit per test standard output Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it...
}else{for(inti =0; i < even.size(); ++i) parts[i].insert(even[i]);intj = k-p;for(inti = even.size(); i < p-1; ++ i) parts[i].insert(odd[j++]),parts[i].insert(odd[j++]);while( j < odd.size()) parts[p-1>=0? p-1: p].insert(odd[j++]); ...
题目//参考了网上的代码注意答案可能超过32位//要达成目标,就是要所有数列a的都比数列b的要小或者等于//然后,要使最小的要和最大的一样大,就要移动(大-小)步,//要使较小的要和较大的一样大,也是要移动(较大-较小)步//然后都加在一起就好了#include#include#include#in