B. Suffix Structure 1. 先判断s去掉一些元素是否能构成t,如果可以就是automaton 判断的方法也很简单,two pointer,相同元素同时++,不相同s的指针++,如果t能全找到,那么s能够去掉元素构成t. bool f(string s, string t) { , j = ; while (i < s.size() && j < t.size()) { if (s[i] == t[j]) { i++; j++; } else { i++; }…

Codeforces #344 Div.2 Interview 题目描述:求两个序列的子序列或操作的和的最大值 solution 签到题 时间复杂度:\(O(n^2)\) Print Check 题目描述:有一个棋盘,对其进行染色,每次染一行或一列,后来的颜色会覆盖原来的颜色,输出最后的棋盘. solution 题解用二维线段树,其实可以不用. 对染色进行离线操作,那么染过的格子.行.列就不用再染了.所以每个格子可以记录四个指针,分别是行前驱,行后继,列前驱列后继,染色时按照这个跳着来染就好了.…

Codeforces #345 Div.1 打CF有助于提高做题的正确率. Watchmen 题目描述:求欧拉距离等于曼哈顿距离的点对个数. solution 签到题,其实就是求有多少对点在同一行或同一列. 时间复杂度:\(O(nlogn)\) Image Preview 题目描述:给定看一张照片的时间,翻页的时间,把图片翻转的时间.一开始屏幕显示第一张照片,可以向左或向右翻,不能跳过还没有看过的图片,方向不对的图片要先翻转再看,看过的不消耗翻转时间与看照片时间,问在一定时间内,最多能看多少张照…

Codeforces Beta Round #27 (Codeforces format, Div. 2) http://codeforces.com/contest/27 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define maxn 1000005 typ…

Codeforces#441 Div.2 四小题 链接 A. Trip For Meal 小熊维尼喜欢吃蜂蜜.他每天要在朋友家享用N次蜂蜜 , 朋友A到B家的距离是 a ,A到C家的距离是b ,B到C家的距离是c  (A- >Rabbit   B- >Owl    C ->Eeyore),他不能连续两顿饭都在同一位朋友家里蹭 他现在位于A的家里, 请问他一天最少要跑多少路. 当然是要找一条最短的路折返跑了啊,是不是很简单. #include<bits/stdc++.h> us…

codeforces #592(Div.2) A Pens and Pencils Tomorrow is a difficult day for Polycarp: he has to attend a lectures and b practical classes at the university! Since Polycarp is a diligent student, he is going to attend all of them. While preparing for th…

codeforces #578(Div.2) A. Hotelier Amugae has a hotel consisting of 1010 rooms. The rooms are numbered from 00 to 99 from left to right. The hotel has two entrances - one from the left end, and another from the right end. When a customer arrives to t…

codeforces #577(Div.2) A  Important Exam A class of students wrote a multiple-choice test. There are nn students in the class. The test had mm questions, each of them had 55 possible answers (A, B, C, D or E). There is exactly one correct answer for…

http://codeforces.com/contest/599/problem/D 题意:给出总的方格数x,问有多少种不同尺寸的矩形满足题意,输出方案数和长宽(3,5和5,3算两种) 思路:比赛的时候gg了..其实稍微在纸上推一下.就会得到对于n,m的矩形,一共会有-n*n*n+3*n*n*m+n+3*n*m的方格.数量级是n3. 我们可以实际跑一遍.发现对于x1E18的数量级,n不会超过1442550,1E6,可以搞. 需要注意的是,一个是会爆int,所以记得用long long 另一个是…

转载请注明出处:viewmode=contents" target="_blank">http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://codeforces.com/contest/448/problem/D -----------------------------------------------------------------------------------------------…

题目链接:http://codeforces.com/contest/448/problem/B ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943/ma…

解题报告 四种情况相应以下四组数据. 给两字符串,推断第一个字符串是怎么变到第二个字符串. automaton 去掉随意字符后成功转换 array 改变随意两字符后成功转换 再者是两个都有和两个都没有 #include <iostream> #include <cstdio> #include <cstring> #include <stdlib.h> #include <algorithm> #include <cmath> usi…

Problem A: A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present - a new glass cup…

A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present - a new glass cupboard with…

 D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isn't just charming, he also is very smart. While some of us were learning the multiplication t…

主题链接:http://codeforces.com/contest/448/problem/D 思路:用二分法 code: #include<cstdio> #include<cmath> #include<iostream> using namespace std; __int64 n,m,k; __int64 f(__int64 x) { __int64 res=0; for(__int64 i=1;i<=n;i++) { __int64 minn=min(…

解题报告 意思就是说有n行柜子,放奖杯和奖牌.要求每行柜子要么全是奖杯要么全是奖牌,并且奖杯每行最多5个,奖牌最多10个. 直接把奖杯奖牌各自累加,分别出5和10,向上取整和N比較 #include <iostream> #include <cstdio> #include <cstring> #include <stdlib.h> #include <algorithm> #include <cmath> using namespa…

D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isn't just charming, he also is very smart. While some of us were learning the multiplication tabl…

题目连接:http://codeforces.com/contest/448 A:给你一些奖杯与奖牌让你推断能不能合法的放在给定的架子上.假设能够就是YES否则就是NO. <span style="font-size:18px;">#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <ioman…

题目链接:http://codeforces.com/problemset/problem/448/C C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. B…

E. Divisors time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isn't just friendly, he also is a rigorous coder. Let's define function f(a), where a is a sequence of intege…

A - Rewards 水题,把a累加,然后向上取整(double)a/5,把b累加,然后向上取整(double)b/10,然后判断a+b是不是大于n即可 #include <iostream> #include <vector> #include <algorithm> #include <cmath> using namespace std; int main(){ double a1,a2,a3; double b1,b2,b3; int n; cin…

C题, #include<cstdio> #include<cstring> #include<algorithm> #define maxn 5005 using namespace std; int num[maxn]; int rmq(int l,int r) { <<,tmp=l; for(int i=l;i<=r;i++) { if(ans>num[i]) { ans=num[i]; tmp=i; } } return tmp; } i…

对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespace std; ],b[]; int main(void) { int n; ; ,sum2 = ; ;i<=;++i){ scanf("%d",&a[i]); sum1 += a[i]; } ;i<=;++i){ scanf("%d",&b[…

解题报告 给篱笆上色,要求步骤最少,篱笆怎么上色应该懂吧,.,刷子能够在横着和竖着刷,不能跳着刷,,, 假设是竖着刷,应当是篱笆的条数,横着刷的话.就是刷完最短木板的长度,再接着考虑没有刷的木板,,. 递归调用,,. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define inf 999999999999999 using namespace…

这次CF状态之悲剧,比赛就别提了.后来应该好好总结. A题:某个细节没考虑到,导致T了 代码: #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> using namespace std; int a[4],b[4],n; int main() { while(scanf("%d%d%d",&…

C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his old fence his…

题目链接 题意: n*m的一个乘法表,从小到大排序后,输出第k个数  (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m) 分析: 对于k之前的数,排名小于k:k之后的数大于,那么就能够採用二分. LL n, m, k; LL fun(LL goal) { LL t = 0, ret = 0; while (++t <= m) { ret += min(n, goal / t); } return ret; } LL bin(LL L, LL R, LL goal) { LL M, V…

Description Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team. At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), s and t. You n…

题意就是将第一个字符串转化为第二个字符串,支持两个操作.一个是删除,一个是更换字符位置. 简单的字符串操作!. AC代码例如以下: #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define M 50010 #define inf 100000000 using namespace std; char a[1005],b[1005]; int la,lb;…