题目链接. 分析: #include <cstdio> #include <iostream> #include <map> #include <cstring> using namespace std; typedef unsigned long long LL; int main() { LL n, m; while(cin >> n >> m) { && m == ) break; LL s = n+m; if(…

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 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…

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…

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…

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 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: 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…

题意: 从左下角移动到右上角.每次只能向上或者向右移动一格.问移动的轨迹形成的右半边图形有多少种 题解: 注意,这个图形就根本不会重复,那就是n*m的图形,向上移动n次,向右移动m次. 从左下角移动到右上角的过程就是n个"上",m个"右"的组合的形式,有多少种移动方式,那就是 C((n+m),n)或者C((n+m),m) C((n+m),n)意思就是从n+m个位置上挑选出来n个位置,这n个位置要向上走,那么剩下m个位置肯定是向右走咯 另外 无符号整型的输入输出用&q…

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…