Description Given a string of 0's and 1's up to 1000000 characters long and indices i and j, you are to answer a question whether all characters between position min(i,j) and position max(i,j) (inclusive) are the same. Input There are multiple cases…

Problem N Zeros and Ones Input: standard input Output: standard output Time Limit: 2 seconds Memory Limit: 32 MB Given a string of 0's and 1's up to 1000000 characters long and indices i and j, you are to answer a question whether all characters betw…

https://vjudge.net/problem/UVA-12063 题意: 统计n为二进制数中,0和1相等且值为m的倍数的数有多少个 dp[i][j][k] 前i位二进制 有j个1 值模m等于k  的数的个数 最高位强制填1,所以实际只需要dp n-1位 #include<cstdio> #include<cstring> using namespace std; ][][]; int main() { int T,n,m; long long ans; scanf(&quo…

题意:给你n.k,问你有多少个n为二进制的数(无前导零)的0与1一样多,且是k的倍数 题解:对于每个k都计算一次dp,dp[i][j][kk][l]表示i位有j个1模k等于kk且第一位为l(0/1) 再次预处理mod[i][j]表示1的i次方模j等于几,具体看代码注释 import java.util.Scanner; public class Main{ static int Maxn=65; static int Maxk=101; //前i个数有j个1模给定的值余k且第一位为1或者0的总个…

#include<bits/stdc++.h> #include<stdio.h> #include<iostream> #include<cmath> #include<math.h> #include<queue> #include<set> #include<map> #include<iomanip> #include<algorithm> #include<stack&g…

1. [HNOI2001] 产品加工 一道简单的背包,然而我还是写了很久QAQ 时间范围是都小于5 显然考虑一维背包,dp[i]表示目前A消耗了i的最小B消耗 注意 if(b[i]) dp[j]=dp[j]+b[i]; else dp[j]=1e9+7; 可以用B则直接转移,否则要把上一次的这个状态设为正无穷,只能用后两个转移. //Twenty #include<cstdio> #include<cstdlib> #include<iostream> #include…

题意: 找出长度为n.0和1个数相等.没有前导0且为k的倍数的二进制数的个数. 分析: 这道题要用动态规划来做. 设dp(zeros, ones, mod)为有zeros个0,ones个1,除以k的余数为mod的二进制数的个数,则状态转移方程为: dp(zeros + 1, ones, (mod>>1) % k) += dp(zeros, ones, mod) dp(zeros, ones + 1, ((mod>>1)+1) % k) += dp(zeros, ones, mod)…

传送门 一开始在vjudge上看到这题时,标的来源是CSU 1120,第八届湖南省赛D题“平方根大搜索”.今天交题时CSU突然跪了,后来查了一下看哪家OJ还挂了这道题,竟然发现这题是出自UVA的,而且用的原题. ------------------------------------------------------------------------------------------------------------------ time limit 5s In binary, the…

十五 Twenty Questions Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1252 Appoint description:  System Crawler  (2015-08-25) Description   Consider a closed world and a set of features that are defined…

这道题目甚长, 代码也是甚长, 但是思路却不是太难.然而有好多代码实现的细节, 确是十分的巧妙. 对代码阅读能力, 代码理解能力, 代码实现能力, 代码实现技巧, DFS方法都大有裨益, 敬请有兴趣者耐心细读.(也许由于博主太弱, 才有此等感觉). 题目: UVa 1103 In order to understand early civilizations, archaeologists often study texts written in  ancient languages. One…