A. Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n cards (n is even) in the deck. Each card has a positive integer written on it. n / 2 people will play new card game. At…

直接看示例输入输出+提示 1. 统计所有数的和 sum,然后求 sum/(n/2) 的到一半数的平均值 6 1 5 7 4 4 3 ->1+5+7+4+4+3=24  分成3组 每组值为8 int sumavg=sum/(n/2)=8 2. int z=sumavg-x[i] if(x[j]==z&&!dp[i]&&!dp[j]) //如果当前数为该输出的数,且i,j没有输出过,则输出 i,j (数组下标), 且dp[i]=dp[j]=1; AC代码可以解释解题答案…

题目链接: A. Cards 题意: 问两个数的和相同,怎么组合; AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j…

A. Bear and Five Cards 题目连接: http://www.codeforces.com/contest/680/problem/A Description A little bear Limak plays a game. He has five cards. There is one number written on each card. Each number is a positive integer. Limak can discard (throw out) s…

C. Soldier and Cards Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/546/problem/C Description Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are diff…

题目链接: A. Bear and Five Cards //#include <bits/stdc++.h> #include <vector> #include <iostream> #include <queue> #include <cmath> #include <map> #include <cstring> #include <algorithm> #include <cstdio>…

B. Cards time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can…

最近在使用codeblock,所以就先刷一些水题上上手 使用codeblock遇到的问题 1.无法进行编译-------从setting中的编译器设置中配置编译器 2.建立cpp后无法调试------只建立源文件无法调试,需要建立一个工程后才能调试 3.设置断点后,调试不会停止------开启-g模式且工程要建立在一个没有中文名的文件夹下 4.调试中如何查看变量------打开debug中的watch,右键编辑界面的变量可以选择添加变量 水题来源---codeforces(hzwer神犇刷的水题…

题目链接:http://codeforces.com/problemset/problem/705/B 题意略解: 两个人玩游戏,解数字,一个数字可以被分成两个不同或相同的数字 (3可以解成 1 2),最小1不能再解 示例1: 第一把 当前1号玩家开始解数字 1,不能解, 则2号赢 输出2 第二把 当前数字为 1 2, 1号玩家直接对 2 进行解 解成 1 1,数字为 1 1 1,2号玩家无解,1号赢 第三把 当前数字为 1 2 3,一号玩家先对 2 解,解成 1 1,数字为 1 1 1 3,2…

题目链接:http://codeforces.com/problemset/problem/327/B 这道题目虽然超级简单,但是当初我还真的没有想出来做法,囧,看完别人的代码恍然大悟. #include <cstdio> #include <cstdlib> #include <cmath> int main(void) { int n; scanf("%d", &n); ; i <= n+n; ++i) { printf("…