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…

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…

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…

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…

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…

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…

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…