Paths on a Grid Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 23270   Accepted: 5735 Description Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastere…

Paths on a Grid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 23008 Accepted: 5683 Description Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered ye…

Paths on a Grid DescriptionImagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he's explaining that (a+b)2=a2+2ab+b2). So you decide to waste…

poj1942 Paths on a Grid 题意:给定一个长m高n$(n,m \in unsigned 32-bit)$的矩形,问有几种走法.$n=m=0$时终止. 显然的$C(m+n,n)$ 但是没有取模,n,m的范围又在unsigned int 范围内 于是有一种神奇的方法↓↓ typedef unsigned us;us C(us a,us b){//C(a,b) double cnt=1.0; while(b) cnt*=(double)(a--)/(double)(b--); re…

Paths on a Grid Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21297   Accepted: 5212 Description Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastere…

Paths on a Grid Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 22918   Accepted: 5651 Description Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastere…

Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he's explaining that (a+b) 2=a 2+2ab+b 2). So you decide to waste your time with drawing…

给定一个矩形网格的长m和高n,其中m和n都是unsigned int32类型,一格代表一个单位,就是一步,求从左下角到右上角有多少种走法,每步只能向上或者向右走. //注意循环的时候,要循环小的数,否则会超时 #include<cstdio> #include<iostream> #define LL long long using namespace std; int main() { ) { LL a,b,ans=; cin>>a>>b; &&a…

处理阶乘有三种办法:(1)传统意义上的直接递归,n的规模最多到20+,太小了,在本题不适用,而且非常慢(2)稍快一点的算法,就是利用log()化乘为加,n的规模虽然扩展到1000+,但是由于要用三重循环,一旦n规模变得更大,耗时就会非常之严重,时间复杂度达到O(n*m*(n-m)),本题规定了n,m用unsigned int32类型,就是说n,m的规模达到了21E以上,铁定TLE的.而且就算抛开时间不算,还存在一个致命的问题,就是精度损失随着n的增加会变得非常严重.因为n有多大,就要进行n次对数…

// n*m 的格子 从左下角走到右上角的种数// 相当于从 n+m 的步数中选 m 步往上走// C(n+m,m) #include <iostream> #include <string> #include<sstream> #include <cmath> #include <map> #include <stdio.h> #include <string.h> #include <algorithm>…