官方英文题解:http://codeforces.com/blog/entry/19237 Problem A: 题目大意: 给出内角和均为120°的六边形的六条边长(均为正整数),求最多能划分成多少个边长为1的正三角形. 题解: 把六边形补全变成一个正三角形,然后减去三个角的正三角形即可. Problem B: 题目大意: 给出长度相等的两个串AB,定义两个串相等 当且仅当  A=B  或者  当长度为偶数时,A[1...n/2]=B[1...n/2]  && A[n/2+1...n]=…

Gerald and Giant Chess Problem's Link: http://codeforces.com/contest/559/problem/C Mean: 一个n*m的网格,让你从左上角走到右下角,有一些点不能经过,问你有多少种方法. analyse: BZOJ上的原题. 首先把坏点和终点以x坐标为第一键值,y坐标为第二键值排序 . 令fi表示从原点不经过任何坏点走到第i个点的个数,那么有DP方程: fi=Cxixi+yi−∑(xj<=xi,yj<=yi)C(xi−xj)…

Equivalent Strings Problem's Link: http://codeforces.com/contest/559/problem/B Mean: 给定两个等长串s1,s2,判断是否等价. 等价的含义为: 若长度为奇数,则必须是相同串. 若长度是偶数,则将两串都均分成长度为原串一半的两个子串l1,r1和l2,r2,其中l1和l2等价且r1和r2等价,或者l1和r2等价且l2和r1等价. analyse: 直接按照题意模拟写个递归分治就行.比赛的时候总觉得这样暴力写会TLE,…

Gerald's Hexagon Problem's Link: http://codeforces.com/contest/559/problem/A Mean: 按顺时针顺序给出一个六边形的各边长(且保证每个内角都是120度),求能够分解成多少个边长为1的小正三角形. analyse: 由于每个内角都是120度,那么把三条边延长相交,一定能够得到一个正三角形. 求出正三角形的面积S1和补上的小三角形的面积S2,则answer=S1-S2. 这里不是真正意义上求正三角形的面积,而是直接求内部可…

B. Gerald is into Art Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/560/B Description Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought…

D. Equivalent Strings Time Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/problem/B Description Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are…

C. Gerald's Hexagon Time Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/problem/A Description Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he me…

A. Currency System in Geraldion Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/560/problem/A Description A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But th…

题目链接:http://codeforces.com/contest/560/problem/E 给你一个n*m的网格,有k个坏点,问你从(1,1)到(n,m)不经过坏点有多少条路径. 先把这些坏点排序一下. dp[i]表示从(1,1)到第i个坏点且不经过其他坏点的路径数目. dp[i] = Lucas(x[i], y[i]) - sum(dp[j]*Lucas(x[i]-x[j], y[i]-x[j])) , x[j] <= x[i] && y[j] <= y[i] //到i…

B. Equivalent Strings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/problem/B Description Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are…

A. Gerald's Hexagon Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/problem/A Description Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he m…

B. Gerald is into Art Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/560/problem/B Description Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought a…

A. Currency System in Geraldion Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/560/problem/A Description A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But t…

大半年没有打Codeforces , 昨天开始恢复打Codeforces, 简直是, 欲语泪先流啊. 手残到爆的写错了范围, 手残的数漏了条件, 简直不能直视, 最坑爹的是, E题没时间写代码了. 题目链接 Problem_A: 题意: 给n个数, 每个数可以用无限次, 求用这些数的和表示不出来的最小的正整数, 没有则输出 -1. 思路: 如果这n个数里面有1, 那么一定可以表示所有数, 没有1的话, 最小的正整数就是1 代码: 1 #include <cmath> 2 #include &l…

[CF简单介绍] 提交链接:http://codeforces.com/contest/560/problem/C 题面: C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Gerald got a very curious hexagon for his birthday. The bo…

C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to .…

C. Gerald and Giant Chess Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/problem/C Description Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . The…

[CF简单介绍] 提交链接:http://codeforces.com/contest/560/problem/B 题面: B. Gerald is into Art time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Gerald bought two very rare paintings at the Sotheby's a…

http://codeforces.com/contest/559/problem/A 题目大意:按顺序给出一个各内角均为120°的六边形的六条边长,求该六边形能分解成多少个边长为1的单位三角形. 解: 性质1:边长为n的正三角形能够划分成n*n个边长为1的正三角形. 绘图找规律 性质2:延长各边总能找到一个大的正三角形.而且所求等于大三角形减去三个补出来的三个三角形面积 收获: 以后先找规律,看能不能找出一些特征即使不会证明 其次,总的减去部分化为所求假设想求的难以直接求 #include <…

[CF简单介绍] 提交链接:http://codeforces.com/contest/560/problem/A 题面: A. Currency System in Geraldion time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A magic island Geraldion, where Gerald lives,…

A http://codeforces.com/contest/560/problem/A 推断给出的数能否组成全部自然数. 水题 int a[1010]; bool b[1000010]; int main() { int n; while (scanf("%d", &n) != EOF) { memset(b,false,sizeof(b)); for (int i = 1; i <= n; i++) { scanf("%d", &a[i]…

这场CF又掉分了... 这题题意大概就给一个h*w的棋盘,中间有一些黑格子不能走,问只能向右或者向下走的情况下,从左上到右下有多少种方案. 开个sum数组,sum[i]表示走到第i个黑点但是不经过其他黑点的方案数. 式子是sum[i]=c(x[i]+y[i],x[i])-Σ(sum[j]*c(x[i]-x[j]+y[i]-y[j],x[i]-x[j])). c(x+y,x)表示从格子(1,1)到(x,y)的方案数(没有黑点). 因此每个点按x[i]+y[i]的值排个序,然后n^2弄一下他们的拓扑…

C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to .…

A. Currency System in Geraldion time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several valu…

感觉题意不太好懂 = =# 给两个字符串 问是否等价等价的定义(满足其中一个条件):1.两个字符串相等 2.字符串均分成两个子串,子串分别等价 因为超时加了ok函数剪枝,93ms过的. #include <iostream> #include <cstring> #define clr(x,c) memset(x,c,sizeof(x)) using namespace std; const int N = 200005; char s[N], t[N]; int sc[30],…

题目链接 题意:  有一个六边形,给你6条边的长度(顺时针给出).每条边都是整数,问你它能够被切割成几个单位长度的正三角形  (题目保证给出的数据能够被切割) 思路: 六边形能够被切割成两种情况: ①被分成上下两个等腰梯形 ②被分成等腰梯形-平行四边形-等腰梯形 事实上这两种能够统为一种,由于当另外一种平行四边形的一对平行边长为0的时间就变成了第一种. 给这六条边标号 六边形旋转到某个状态的时候. 一定会形成等腰梯形-平行四边形-等腰梯形的状态(或仅仅有两个等腰梯形) 上下两个梯形的腰各自是 m…

题意:定义了字符串的相等,问两串是否相等. 卡了时间,空间,不能新建字符串,否则会卡. #pragma comment(linker,"/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<vector> #include<cmath> #inclu…

题意:面积是sqrt(3)/4的多少倍? 做延长线 #pragma comment(linker,"/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<vector> #include<cmath> #include<queue> #i…

A题,超级大水题,根据有没有1输出-1和1就行了.我沙茶,把%d写成了%n. B题,也水,两个矩形的长和宽分别加一下,剩下的两个取大的那个,看看是否框得下. C题,其实也很简单,题目保证了小三角形是正三角形,一个正三角的面积=l*l*(1/2)*cos(30),由于只要算三角形个数,把六边形扩成一个大三角,剪掉三个小三角,除一下系数就没了.就变成了平方相减. D题,根据定义递归.然后注意奇数就行了.我沙茶,没加第一种判断dfs(a+len,len,b+len) && dfs(a,len,b…

D. Equivalent Strings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal l…