题意: n道题,每道题有ai和bi,完成这道题需要先完成若干道题,完成这道题可以得到分数t*ai+bi,其中t是时间 1s, n<=20 思路: 由n的范围状压,状态最多1e6 然后dfs,注意代码中dfs里的剪枝, 对一个状态statu,因为贪心的取最大值就行,所以及时剪枝 代码: 当时写不出来真是菜的活该 #include<iostream> #include<cstdio> #include<algorithm> #include<cmath>…

题目链接: https://nanti.jisuanke.com/t/30994 Dlsj is competing in a contest with n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems. However, he can submit ii-th problem if and only if he has submitted (and passed, of c…

2018ICPC南京网络赛 A. An Olympian Math Problem 题目描述:求\(\sum_{i=1}^{n} i\times i! \%n\) solution \[(n-1) \times (n-1)! \% n= (n-2)!(n^2-2n+1) \%n =(n-2)!\] \[(n-2+1)\times (n-2)! \% n= (n-3)!(n^2-3n+2) \%n =(n-3)! \times 2\] 以此类推,最终只剩下\(n-1\) 时间复杂度:\(O(1)\…

题目链接:https://nanti.jisuanke.com/t/30994 样例输入: 5 5 6 0 4 5 1 1 3 4 1 2 2 3 1 3 1 2 1 4 样例输出: 55 样例输入: 1 -100 0 0 样例输出: 0 题解: 把n道题目做了或者没做作为状态,裸的状压DP. 其中当前的时间 t,就是当前做了的题目数量加上1. AC代码: #include<bits/stdc++.h> using namespace std; typedef long long ll; co…

题意: 有向图,可以把k条路的长度变为0,求1到n的最短路 思路: 将图复制k份,一共k+1层图,对于每一条i→j,都连一条低层的i→高层的j,并且权值为0 即对每一对<i,j,w>,都加边<i,j,w>,<i+n,j+n,w>,<i+2n,j+2n,w>,....,<i+kn,j+kn,w> 同时加“楼梯”<i,j+n,0>,<i+n,j+2n,0>,...,<i+(k-1)n, j+kn> 然后跑一个1~(…

Walk Through Squares Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 200    Accepted Submission(s): 57 Problem Description   On the beaming day of 60th anniversary of NJUST, as a military colleg…

Divide Groups Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 423    Accepted Submission(s): 161 Problem Description   This year is the 60th anniversary of NJUST, and to make the celebration mor…

Count The Pairs Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 277    Accepted Submission(s): 150 Problem Description   With the 60th anniversary celebration of Nanjing University of Science…

2019ICPC南京网络赛A题 The beautiful values of the palace https://nanti.jisuanke.com/t/41298 Here is a square matrix of n * nn∗n, each lattice has its value (nn must be odd), and the center value is n * nn∗n. Its spiral decline along the center of the squar…

南京网络赛自闭现场 https://nanti.jisuanke.com/t/41298 二维偏序经典题型 二维前缀和!!! #include<bits/stdc++.h> using namespace std; #define int long long #define sc(x) scanf("%lld",&x); int T; #define P pair<int,int> #define fi first #define se second #…