HDU 5860 Death Sequence(递推) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5860 Description You may heard of the Joseph Problem, the story comes from a Jewish historian living in 1st century. He and his 40 comrade soldiers were trapped in a cave
Problem Description You may heard of the Joseph Problem, the story comes from a Jewish historian living in 1st century. He and his 40 comrade soldiers were trapped in a cave, the exit of which was blocked by Romans. They chose suicide over capture an
Time limit: 1.0 second Memory limit: 64 MB On the Day of the Flag of Russia a shop-owner decided to decorate the show-window of his shop with textile stripes of white, blue and red colors. He wants to satisfy the following conditions: Stripes of the
1.直接用递归函数计算状态转移方程,效率十分低下,可以考虑用递推方法,其实就是“正着推导,逆着计算” #include<iostream> #include<algorithm> using namespace std; #define maxn 1000+5 int n; int a[maxn][maxn]; int d[maxn][maxn]; int main(){ for(;cin>>n && n;){ memset(d,,sizeof(d));
矩阵乘法,顾名思义矩阵与矩阵相乘, 两矩阵可相乘的前提:第一个矩阵的行与第二个矩阵的列相等 相乘原则: a b * A B = a*A+b*C a*c+b*D c d C D = c*A+d*C c*A+d*C 上代码 struct matrix { ll a[maxn][maxn]; }; matrix matrix_mul(matrix x,matrix y) { matrix temp; ;i<=n;i++) ;j<=n;j++) { tem
递推 1什么是递推?:根据已有节点的值,以及规律推出之后节点的值 2为什么要用递推:简单的解决有规矩事件 3怎么用?: 我们举个经典的例子: 如果1对兔子每月能生1对小兔子,而每对小兔在它出生后的第3个月就可以生1对小兔子,如果从1对初生的小兔子开始,1年后能繁殖多少兔子? def my1(max): a ,b,c ,i= 1,0,0 0 while i<max: c = c+b b = a a = c print a+b+c i+=1 方法:我们可以把兔子分为1个月大的,2个月大的,3个月大的
小兔的棋盘 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9447 Accepted Submission(s): 4879 Problem Description 小兔的叔叔从外面旅游回来给她带来了一个礼物,小兔高兴地跑回自己的房间,拆开一看是一个棋盘,小兔有所失望.不过没过几天发现了棋盘的好玩之处.从起点(0,0)走到终点(n
