Gridland(规律)】的更多相关文章

Gridland Time Limit: 2 Seconds      Memory Limit: 65536 KB BackgroundFor years, computer scientists have been trying to find efficient solutions to different computing problems. For some of them efficient algorithms are already available, these are t…
题目链接 多写几个案例,根据数据的奇偶性,就能找到规律了 #include<stdio.h> int main() { int t,n,m; double ans; scanf("%d",&t); ;i<=t;i++) { scanf("%d%d",&n,&m); ans=n*m*1.0; ==&&m%==) ans+=0.41; printf("Scenario #%d:\n%.2lf\n\n&q…
分析:给出一个矩阵.问最短从一个点经过全部点以此回到起点的长度是多少.绘图非常好理解.先画3*4.3*3.4*4的点阵图案.试着在上面用最短路走一走,能够发现当矩形点阵的长宽都是奇数时,最短路中必然有一条斜线:而仅仅要长或宽有一个是偶数就能够通过直线来完毕最短路经.因此仅仅需推断一下两边的奇偶情况就能求最短路径了. #include<iostream> #include<cmath> using namespace std; int main() { int T,t=0,m,n;…
Problem Description For years, computer scientists have been trying to find efficient solutions to different computing problems. For some of them efficient algorithms are already available, these are the “easy” problems like sorting, evaluating a pol…
最重要的是找规律. 下面是引用 http://blog.sina.com.cn/s/blog_4dc813b20100snyv.html 的讲解: 做这题时,千万不要被那个图给吓着了,其实这题就是道简单的数学题. 首先看当m或n中有一个为2的情况,显然,只需要算周长就OK了.即(m+n-)*,考虑到至少其中一个为2,所以答案为2 *m或2*n,亦即m*n.注意这里保证了其中一个数位偶数. 当m,n≥3时,考虑至少其中一个为偶数的情况,显然,这种情况很简单,可以得出,结果为m*n,又可以和上面这种…
Happy 2004 题意:s为2004^x的因子和,求s%29.     (题于文末) 知识点: 素因子分解:n = p1 ^ e1 * p2 ^ e2 *..........*pn ^ en 因子和:    Sum=(p1^0+p1^1-.p1^e1)*(p2^0+p2^1-p2^e2)--(pn^0+-pn^en) =; 积性函数:s(xy)=s(x)*s(y)    (比如:幂函数,因子和,欧拉函数,莫比乌斯函数) 对于正整数n的一个算术函数 f(n),若f(1)=1,且当a,b互质时f…
传送门 Description Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a…
//POJ3377 //DP解法-解有规律的最短路问题 //Time:1157Ms Memory:12440K #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAXN 1000005 typedef long long LL; int n; int dp[MAXN][3]; int sr, st, er, ed; int main() { //freo…
A Simple Nim Problem Description   Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed)…
--利用sqlserver来运算斐波那契规律 --利用事物与存储过程 declare @number intdeclare @A intdeclare @B intdeclare @C int set @A=1 set @B=2 set @Number=3 select @C=@A+@B while(@Number<60) begin    set @C=@A+@B  if(@@ERROR<>0)  goto errorhandler  print N'第'+convert(varcha…