我感觉这道题挺神的~ 假设 $a[i]=b[i]$,那么我们可以将 $a$ 降序排序,然后你发现只要你按照 $1,3,5......n$ 这么取一定是合法的. 而我们发现 $2$ 比取 $3$ 优,取 $4$ 还比取 $5$ 优. 所以,我们可以这样: 强制性取第一个元素,然后其余 $\frac{n}{2}$ 个元素每相邻两个依次考虑,不论拿哪个都是合法的. 这样做有什么好处呢?由于 $a$ 可以这样随便拿,于是每一次就取更大的 $b$ 就好了. 所以,我们按照 $a$ 从大到小排一个序,然后依…

CF798D Mike and distribution 洛谷评测传送门 题目描述 Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive in…

Mike and distribution 题目链接:http://codeforces.com/problemset/problem/798/D 数据范围:略. 题解: 太难了吧这个题..... 这种贪心根本就不咋会....接下来刷一段时间Atcoder看看好了..... 就是想到先把所有的对按照$a$排序. 然后刨除第一个外,相邻的两个分组. 第一个数单独一组,剩下的两两一组选$b$值较大的那对即可. 证明: 首先,因为我们按照$a$排序,所以上一组的选取的$a$一定不比当前组剩下的$a$小…

Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a1, a2, ..., an] and B = [b1, …

题目链接 Mike and distribution 题目意思很简单,给出$a_{i}$和$b_{i}$,我们需要在这$n$个数中挑选最多$n/2+1$个,使得挑选出来的 $p_{1}$,$p_{2}$,$p_{3}$,...,$p_{m}$满足 $a_{p1}+a_{p2}+a_{p3}+...+a_{p_{m}}>a_{1}+a_{2}+a_{3}+...+a_{n}$ $b_{p1}+b_{p2}+b_{p3}+...+b_{p_{m}}>b_{1}+b_{2}+b_{3}+...+b_…

D. Mike and distribution time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it…

D. Mike and distribution time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it…

http://codeforces.com/contest/798/problem/D D. Mike and distribution time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mike has always been thinking about the harshness of social inequality.…

/* CF410div2 D. Mike and distribution http://codeforces.com/contest/798/problem/D 构造 题意:给出两个数列a,b,求选出n/2+1个数对,使得其和的二倍大于各自的数列 思路:对数列a进行排序,因为可以选一半加1个,所以最大的那个我们选出来 然后在剩下的数列中,每隔两个选则b中较大的, 这样可以保证选出的在b中满足条件,并且在a中也满足条件 然而....我他喵的居然忘了读入b数列!!!! */ #include <c…

http://codeforces.com/contest/798/problem/D http://blog.csdn.net/yasola/article/details/70477816 对于二维的贪心我们可以先让它变成其中一维有序,这样只需要重点考虑另一维,就会简单很多. 首先,对于题目要求的选择元素之和两倍大与所有元素之和,我们可以转化为选择元素之和大于剩下的.然后我们可以将下标按照a从大到小排序.然后选择第一个,之后每两个一组,选择b大的一个,如果n是偶数再选择最后一个. 至于这样写…