摘要: 题意 给出n行m列的迷宫0可走1不可走,有两个操作,操作1变换点(a,b)的值,操作2查询(1,a)到(n,b)的方案数 题解 设$F[i][j]$为第i 1行到达第i行第j列的方案数,若点$(i,j)$上下为0的可延伸范围为$(l,r)$,则$F[i][j] = \sum_{k=l}^r F[i 阅读全文
posted @ 2019-07-26 11:37 奔跑的蜗new 阅读(190) 评论(0) 推荐(0)
摘要: 题目连接: https://ac.nowcoder.com/acm/contest/883/H Description There are always some problems that seem simple but is difficult to solve. ZYB got N disti 阅读全文
posted @ 2019-07-26 02:52 奔跑的蜗new 阅读(275) 评论(0) 推荐(0)
摘要: 题目连接: https://ac.nowcoder.com/acm/contest/882/D Description Given a vertex weighted graph with N vertices, find out the K th minimum weighted clique. 阅读全文
posted @ 2019-07-25 20:01 奔跑的蜗new 阅读(171) 评论(0) 推荐(0)
摘要: 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=6601 Description N sticks are arranged in a row, and their lengths are a1,a2,...,aN. There are Q query 阅读全文
posted @ 2019-07-25 00:53 奔跑的蜗new 阅读(398) 评论(0) 推荐(0)
摘要: 题目连接: https://vjudge.net/problem/SPOJ SUBLEX Description Little Daniel loves to play with strings! He always finds different ways to have fun with str 阅读全文
posted @ 2019-07-23 11:14 奔跑的蜗new 阅读(165) 评论(0) 推荐(0)
摘要: 题目连接: https://vjudge.net/problem/SPOJ NSUBSTR Description You are given a string S which consists of 250000 lowercase latin letters at most. We define 阅读全文
posted @ 2019-07-19 19:04 奔跑的蜗new 阅读(176) 评论(0) 推荐(0)
摘要: 题目连接: https://vjudge.net/problem/SPOJ LCS2 Description A string is finite sequence of characters over a non empty finite set Σ. In this problem, Σ is 阅读全文
posted @ 2019-07-19 15:55 奔跑的蜗new 阅读(161) 评论(0) 推荐(0)
摘要: 题目连接: https://vjudge.net/problem/SPOJ LCS2 Description A string is finite sequence of characters over a non empty finite set Σ. In this problem, Σ is 阅读全文
posted @ 2019-07-19 15:26 奔跑的蜗new 阅读(187) 评论(0) 推荐(0)
摘要: 题目连接: https://vjudge.net/problem/HDU 6053 Description You are given an array A , and Zhu wants to know there are how many different array B satisfy th 阅读全文
posted @ 2019-07-17 10:20 奔跑的蜗new 阅读(146) 评论(0) 推荐(0)
摘要: #include using namespace std; const int mx = 1e5+5; typedef long long ll; bool vis[mx]; int sum[mx], prim[mx], mu[mx], cnt = 0; void get_mu(){ mu[1] = 1; for(int i = 2; i< mx; i++){ ... 阅读全文
posted @ 2019-07-16 19:56 奔跑的蜗new 阅读(129) 评论(0) 推荐(0)