题意:打怪兽.可增加自己的属性,怎样在能打倒怪兽的情况下花费最少? 这题关键要找好二分的量.一开始我觉得,只要攻击到101,防御到100,就能必胜,于是我对自己的三个属性的和二分(0到201),内部三层循环(最多到不了200*200*200).1秒内能过.不过发现如果生命值很便宜,防御很贵的话,买生命值合算.10100点生命值就能必赢,于是上界调为10100,超时. 后来就想,二分攻击(记为i)和防御(记为j)的和mid,内部二重循环列出i+j=mid的所有情况.再单独二分生命值k,如果ijk的…

A. Fight the Monster Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/487/problem/A Description A monster is attacking the Cyberland! Master Yang, a braver, is going to beat the monster. Yang and the monster each have 3 attr…

题目链接:http://codeforces.com/contest/488 A. Giga Tower Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from  - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor  - 1 and…

Fight the Monster time limit per test             1 second                                   memory limit per test       256 megabytes                                   input                               standard input                               …

B. Strip Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/487/problem/B Description Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some pieces (p…

B. Candy Boxes Problem's Link:   http://codeforces.com/contest/488/problem/B Mean: T题目意思很简单,不解释. analyse: 这道题还是很有意思的,需要考虑到各种情况才能AC. 解这个题目之前,首先要推出两条式子 x4=3x1 4x1=x2+x3 然后就是分类讨论,枚举各种情况就可. Time complexity: O(1) Source code:  // Memory Time // 1347K 0MS…

A 这么简单的题直接贴代码好了. #include <cstdio> #include <cmath> using namespace std; bool islucky(int a) { a = abs(a); while(a) { == ) return true; a /= ; } return false; } int main(void) { ; scanf("%d", &a); while(!islucky(a + b)) b++; prin…

A A monster is attacking the Cyberland! Master Yang, a braver, is going to beat the monster. Yang and the monster each have 3 attributes: hitpoints (HP), offensive power (ATK) and defensive power (DEF). During the battle, every second the monster's H…

D. Strip time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some p…

D - Conveyor Belts 思路:分块dp, 对于修改将对应的块再dp一次. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; ;…

B - Strip 思路:简单dp,用st表+单调队列维护一下. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; ; ; const int…

Strip time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some piec…

哎,最近弱爆了,,,不过这题还是不错滴~~ 要考虑完整各种情况 8795058                 2014-11-22 06:52:58     njczy2010     B - Candy Boxes             GNU C++     Accepted 31 ms 4 KB 8795016                 2014-11-22 06:48:15     njczy2010     B - Candy Boxes             GNU C+…

题目链接: A. Fight the Monster time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A monster is attacking the Cyberland! Master Yang, a braver, is going to beat the monster. Yang and the monster ea…

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…

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…

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 =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

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…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

A. Watching a movie time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have decided to watch the best moments of some movie. There are two buttons on your player: Watch the current minute…

CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了. A 水题 //#pragma comment(linker, "/STACK:102400000,102400000") #include<cstdio> #include<cmath> #include<iostream> #include<…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…

#include <iostream> #include <string> using namespace std; int main(){ int n; cin >> n; string str; cin >> str; , x = ; ; i < n ; ++ i){ if(str[i] == 'B') cnt+=(x << i); } cout<<cnt<<endl; }   Codeforces Round…

Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一个queen了,问你有多少种方案,能够得到n点 题解 其实只用考虑三种情况,都考虑一下就好了,实在不行就暴力枚举-- 代码 #include<bits/stdc++.h> using namespace std; int n; int main() { scanf("%d",&…

Codeforces Round #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须买q[i]个东西,然后他会送你{0,1,2}个物品,但是送的物品必须比你买的最便宜的物品还便宜,问你最少花多少钱,买完m个物品 题解 显然我选择q[i]最小的去买就好了 代码 #include<bits/stdc++.h> using namespace std; const int maxn =…

Codeforces Round #383 (Div. 2) A. Arpa's hard exam and Mehrdad's naive cheat 题意 求1378^n mod 10 题解 直接快速幂 代码 #include<bits/stdc++.h> using namespace std; long long quickpow(long long m,long long n,long long k) { long long b = 1; while (n > 0) { if…