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Div1A / 2C. Make Nonzero Sum 令最后每个\(a_i\)的系数为\(c_i\)(\(c_i=1/-1\)),发现只要满足\(c_1=1\)(下标从1开始),且c中没有两个-1相连,就一定能找出一种划分方式.那我们先令所有\(c_i\)都为1,再进一步把一些1改成-1.如果全是1时序列的和sum已经是0,那么就已经找到一个答案了.否则我们只会把\(a_i=1/-1\)的位置的系数改成-1,当\(sum>0\)时改\(a_i=1\)的i的系数,否则改\(a_i=-1\)的i…
CodeForces1754 注:所有代码均为场上所书 Technical Support 解析: 题目大意 给定一个只包含大写字母 \(\texttt{Q}\) 和 \(\texttt{A}\) 的字符串,如果字符串里的每一个 \(\texttt{Q}\) 都能与在其之后的 \(\texttt{A}\) 一一对应地匹配,则输出字符串 \(\texttt{Yes}\),否则输出字符串 \(\texttt{No}\).注意,可以有 \(\texttt{A}\) 没有被匹配,但每个 \(\textt…
比赛链接 A 题解 知识点:枚举. 只要一个Q后面有一个A对应即可,从后往前遍历,记录A的数量,遇到Q则数量减一,如果某次Q计数为0则NO. 时间复杂度 \(O(n)\) 空间复杂度 \(O(1)\) 代码 #include <bits/stdc++.h> #define ll long long using namespace std; bool solve() { int n; cin >> n; string s; cin >> s; s = "?&qu…
题目链接 题目大意: \(~~\)给定n个正整数和一个数k,问这n个数的阶乘之和能不能被k的阶乘整除 既:(a\(_{1}\)!+a\(_{2}\)!+a\(_{3}\)!+....+a\(_{n}\)!)\(~\)%\(~\)k!\(~\)==$~$0 题目分析: 我们把连续几个数的阶乘可以分解成如下结果: \(~~~\)假设现在有3个2,4个3: \(~~~\)那我们可以得到sum\(~\)=$~$3*2!+ 4*(3*2!) \(~~~~~~~~~~~~~~~~~~~~~~~~~\)sum…
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…
Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…
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Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…