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…

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)…

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…

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…

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…

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. 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…

[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 .…

[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…

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'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 .…

Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~ The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the follow…

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…

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 <…

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: 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…

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…

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. 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…

[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,…

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]=…

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,…

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…

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…

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]…

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…

A题: 题目地址:Currency System in Geraldion 题意:给出n中货币的面值(每种货币有无数张),要求不能表示出的货币的最小值.若全部面值的都能表示,输出-1. 思路:水题,就是看看有没有面值为1的货币,假设有的话,全部面值的货币都能够通过1的累加得到,假设没有的话.最小的不能表示的就当然是1辣. #include <stdio.h> #include <math.h> #include <string.h> #include <stdli…