题目传送门 /* 题意:n个数字转盘,刚开始每个转盘指向一个数字(0~n-1,逆时针排序),然后每一次转动,奇数的+1,偶数的-1,问多少次使第i个数字转盘指向i-1 构造:先求出使第1个指向0要多少步,按照这个次数之后的能否满足要求 题目读的好累:( */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath>…

B. Case of Fake Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/556/problem/B Description Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famou…

题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0),先从1开始找到已经套好的娃娃层数, 其他是2次操作,还要减去k-1个娃娃是只要套上就可以 详细解释:http://blog.csdn.net/firstlucker/article/details/46671251 */ #include <cstdio> #include <algor…

题目传送门 /* 找规律/贪心:ans = n - 01匹配的总数,水 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; char s[MAXN]; int main(void) //Codeforce…

C. Case of Chocolate Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/C Description Andrewid the Android is a galaxy-known detective. Now he does not investigate any case and is eating chocolate out of boredom. A…

A. Case of the Zeros and Ones Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/556/problem/A Description Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros an…

B. Case of Fugitive Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/B Description Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost ful…

C. String Manipulation 1.0 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/A Description Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of cont…

B. Case of Fugitive time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet a…

题目地址:传送门 这题尽管是DIV1的C. . 可是挺简单的. .仅仅要用线段树分别维护一下横着和竖着的值就能够了,先离散化再维护. 每次查找最大的最小值<=tmp的点,能够直接在线段树里搜,也能够二分去找. 代码例如以下: #include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm> #include <…