Poor Hanamichi Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 743    Accepted Submission(s): 275 Problem Description Hanamichi is taking part in a programming contest, and he is assigned to so…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4956 Poor Hanamichi Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7    Accepted Submission(s): 4 Problem Description Hanamichi is taking part in…

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 354    Accepted Submission(s): 100 Problem Description ZYB has a tree with N nodes,now he wants you to solve the numbers of nodes distanced no m…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 175    Accepted Submission(s): 74 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutat…

BestCoder Sequence Problem Description Mr Potato is a coder. Mr Potato is the BestCoder. One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence. A…

Task schedule Problem Description 有一台机器,而且给你这台机器的工作表.工作表上有n个任务,机器在ti时间运行第i个任务,1秒就可以完毕1个任务. 有m个询问,每一个询问有一个数字q.表示假设在q时间有一个工作表之外的任务请求,请计算何时这个任务才干被运行. 机器总是依照工作表运行,当机器空暇时马上运行工作表之外的任务请求.   Input 输入的第一行包括一个整数T, 表示一共同拥有T组測试数据. 对于每组測试数据: 第一行是两个数字n, m,表示工作表里面有…

基本数学题一道,看错位数,当成大数减做了,而且还把方向看反了.所求为最接近l的值. #include <cstdio> int f(__int64 x) { int i, sum; i = sum = ; while (x) { ) sum -= x%; else sum += x%; ++i; x/=; } return sum; } int main() { __int64 l, r, x; bool flg; int t; scanf("%d", &t); w…

GT and numbers 问题描述 给出两个数NN和MM. NN每次可以乘上一个自己的因数变成新的NN. 求最初的NN到MM至少需要几步. 如果永远也到不了输出-1−1. 输入描述 第一行读入一个数TT表示数据组数. 接下来TT行,每行两个数NN和MM. T\leq1000T≤1000, 1\leq N \leq 10000001≤N≤1000000,1 \leq M \leq 2^{63}1≤M≤2​63​​. 注意M的范围.hack时建议输出最后一行的行末回车;每一行的结尾不要输出空格.…

比赛链接 A题主要是map的使用,比赛的时候问了下队友,下次要记住了 #include<bits/stdc++.h> using namespace std; typedef long long LL; LL T,n; map<string,int> mp1,mp2; int main() { cin>>T; while(T--) { mp1.clear();mp2.clear(); cin>>n; ;i<n;i++) { string str; in…

HDU 5904 - LCIS [ DP ]    BestCoder Round #87 题意: 给定两个序列,求它们的最长公共递增子序列的长度, 并且这个子序列的值是连续的 分析: 状态转移方程式: dp[a[i]] = max(dp[a[i]], dp[a[i]-1] + 1); 发现其实可以简化为 dp[a[i]] = dp[a[i]-1] + 1:因为计算过程中dp[a[i]]不会降低 对两个序列都求一遍,然后取两者最小值的最大值 #include <cstdio> #include…