题目链接 将给出的已经有了的排序, 在前面加上0, 后面加上1e9+1. 然后对相邻的两项判断. 如果相邻两项之间的数的和小于m, 那么全都选上, m减去相应的值. 如果大于m, 那么二分判断最多能选多少个. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <complex>…

题目链接:http://codeforces.com/problemset/problem/659/C 题意: n是已经有的数字,m是可用的最大数字和 要求选自己没有的数字,且这些数字的数字和不能超过m 且要求可选的数字的数目越多越好 输出一种答案即可 解题思路: 刚开始想开一个bool型的1e9的数组,然后判断即可 可是交上去发现内存超限 后来把1e9的数组改成2*1e6的数组即可 具体原因应该从数学数字和方面考虑 具体代码如下: #include<bits/stdc++.h> using…

题目链接: C. Tanya and Toys time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered wi…

C. Tanya and Toys 题目链接http://codeforces.com/contest/659/problem/C Description In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collectio…

C. Tanya and Toys 题目连接: http://www.codeforces.com/contest/659/problem/C Description In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new col…

题目链接:http://codeforces.com/problemset/problem/518/B 题目意思:给出字符串 s 和 t,如果 t 中有跟 s 完全相同的字母,数量等于或者多过 s,就将 s 这个数量加到 YAY! 的答案里,少于的话就加 t 中有的数量:如果 t 中有跟 s 相同的字母但是大小写不对应(例如A-a,z-Z),就加到 WHOOPS 的答案里. 举个例子就很容易明白了.例如 s 和 t 分别为: ncMeXssLHS uwyeMcaFatpInZVdEYpwJQSn…

http://poj.org/problem?id=2318 TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10178   Accepted: 4880 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child…

C - Preparing for the Contest Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 377B Description Soon there will be held the world's largest programming contest, but the testing system sti…

题目链接:Codeforces 484B Maximum Value 题目大意:给定一个序列,找到连个数ai和aj,ai%aj尽量大,而且ai≥aj 解题思路:类似于素数筛选法的方式,每次枚举aj,然后枚举k,每次用二分找到小于k∗aj而且最大的ai,维护答案,过程中加了一些剪枝. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn =…

Description In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1to 109. A toy from the new collection of the i-th type costs i bourles. Tania has managed to collect n …