题目链接:http://lightoj.com/volume_showproblem.php?problem=1102 As I am fond of making easier problems, I discovered a problem. Actually, the problem and k=. There are solutions. They are . . . . . . . . . . . . . . . As I have already told you that I us…

1102 - Problem Makes Problem As I am fond of making easier problems, I discovered a problem. Actually, the problem is 'how can you make n by adding k non-negative integers?' I think a small example will make things clear. Suppose n=4 and k=3. There a…

瞬间移动 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2404    Accepted Submission(s): 1066 Problem Description 有一个无限大的矩形,初始时你在左上角(即第一行第一列),每次你都可以选择一个右下方格子,并瞬移过去(如从下图中的红色格子能直接瞬移到蓝色格子),求到第n 行第m 列的格…

题目大意: 给你一菱形的数字阵,问从最上面走到最下面所能获得的最大值是多少? #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #include<queue> #include<vector> #include<map> using namespace std; typedef lo…

题意: 给你一个数 N , 求分成 K 个数 (可以为 0 ) 的种数: 思路: 类似 在K个抽屉放入 N 个苹果, 不为0, 就是 在 n-1 个空隙中选 m-1个: 为 0, 就可以先在 K 个抽屉一个苹果, 之后类似了: 故答案就是 C(N+K-1, K-1): 数据大, 还控制内存... 按位乘 + 逆元 #include<bits/stdc++.h> using namespace std; typedef int LL; const int maxn = 2000000 + 131…

题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1067 1067 - Combinations Given n different objects, you want to take k of them. How many ways to can do it? For example, say there are 4 items; you want to take 2 of them. So, you can do it 6…

1102 - Problem Makes Problem As I am fond of making easier problems, I discovered a problem. Actually, the problem is 'how can you make n by adding k non-negative integers?' I think a small example will make things clear. Suppose n=4and k=3. There ar…

Description Given n different objects, you want to take k of them. How many ways to can do it? For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways. Take 1, 2 Take 1, 3 Take 1, 4 Take 2, 3 Take 2, 4 Take 3, 4 Input…

标题来源:problem=1406">Light OJ 1406 Assassin`s Creed 意甲冠军:向图 派出最少的人经过全部的城市 而且每一个人不能走别人走过的地方 思路:最少的的人能够走全然图 明显是最小路径覆盖问题 这里可能有环 所以要缩点 可是看例子又发现 一个强连通分量可能要拆分 n最大才15 所以就状态压缩 将全图分成一个个子状态 每一个子状态缩点 求最小路径覆盖 这样就攻克了一个强连通分量拆分的问题 最后状态压缩DP求解最优值 #include <cstdio…

C. Beautiful Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal…