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…

题意:格路问题 没什么难度 难点在于如何快速计算相对较大的组合数 思路:运用手写计算组合数的方式进行计算  如c(8,3) 如果手算就是   8*7*6/(3*2*1)这样可以很快得解出 计算代码为:(精度没问题? 反正能过) u c(u n,u m){ u a=n+m; u b=min(n,m); ; ){ ans*=(1.0*a--)/(1.0*b--); } ans+=0.5;//四舍五入 return u(ans); } AC代码: #include<cstdio> #include&…

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…

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…

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…

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…

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…

Milking Grid POJ - 2185 最小覆盖子串: 最小覆盖子串(串尾多一小段时,用前缀覆盖)长度为n-next[n](n-pre[n]),n为串长. 当n%(n-next[n])==0时,有最小循环节(就是最小覆盖子串). 快照: 我对KMP的一些理解(lyp点拨的):pre[i](或next[i])的实质是串str[1..i]的最长且小于i的“相等前.后缀”分别为str[1..pre[i]](前缀)与str[(i-pre[i]+1)..i](后缀),通俗讲就是:使str[1..i…