10399: F.Turing equation Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 151  Solved: 84 [Submit][Status][Web Board] Description The fight goes on, whether to store  numbers starting with their most significant digit or their least  significant digit…

Turing equation 时间限制: 1 Sec 内存限制: 128 MB 题目描述 The fight goes on, whether to store numbers starting with their most significant digit or their least significant digit. Sometimes this is also called the "Endian War". The battleground dates far bac…

Turing equation Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 152  Solved: 85 [Submit][Status][Web Board] Description The fight goes on, whether to store  numbers starting with their most significant digit or their least  significant digit. Sometim…

Hard to Believe, but True! Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3537   Accepted: 2024 Description The fight goes on, whether to store numbers starting with their most significant digit or their least significant digit. Sometim…

A 海岛争霸 题目:Q次询问,他想知道从岛屿A 到岛屿B 有没有行驶航线,若有的话,所经过的航线,危险程度最小可能是多少. 多源点最短路,用floyd 在松弛更新:g[i][k] < g[i][j] && g[k][j] < g[i][j]时,g[i][j] = max(g[i][k],g[k][j]); #include<bits/stdc++.h> using namespace std; const int maxn = 510; int n,m; int q…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=3333 Description After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new…

B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following proble…

 FZU 2102   Solve equation Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description You are given two positive integers A and B in Base C. For the equation: A=k*B+d We know there always existing many non-nega…

题意: 有1~9数字各有a1, a2, -, a9个, 有无穷多的+和=. 问只用这些数字, 最多能组成多少个不同的等式x+y=z, 其中x,y,z∈[1,9]. 等式中只要有一个数字不一样 就是不一样的 思路: 计算下可以发现, 等式最多只有36个. 然后每个数字i的上界是17-i个 可以预先判掉答案一定是36的, 然后直接暴力搜索每个等式要不要就好了. 注意剪枝即可 ; int a[maxn]; bool flag36; int ans; struct Equation { int x, y…

#对coursera上Andrew Ng老师开的机器学习课程的笔记和心得: #注:此笔记是我自己认为本节课里比较重要.难理解或容易忘记的内容并做了些补充,并非是课堂详细笔记和要点: #标记为<补充>的是我自己加的内容而非课堂内容,参考文献列于文末.博主能力有限,若有错误,恳请指正: #---------------------------------------------------------------------------------# 多元线性回归的模型: #-----------…