CodeForces 758 D Ability To Convert】的更多相关文章

Ability To Convert 题意:给你一个n进制的60位的数,但是由于Alexander只会写0->9,所以他就会用10来表示十而不是A(假设进制>10); 题解:模拟就好了,先走往前走进制的位数的倍数,再判断一下首位是不是0,或者这个数有没有大于等于进制,如果有就不行,就要往后走,走到一个非0开头的点,或者就是走到只有1个0的点,然后算出目前这段区间的答案加一个倍数就好了. 代码: #include<bits/stdc++.h> using namespace std;…
[题目链接]:http://codeforces.com/contest/758/problem/D [题意] 给你一个n进制的数k; 问你它可能的最小的十进制数是多少; [题解] 从右往左; 获取数字; 如果这个数字小于n就一直往左扩大; 尽量大; 这样把尽可能多的数字放在权值的低位; 这样n进制数的位就会尽可能地少; 十进制数也就尽可能地小了; 但是有前导0的情况; 这个判断前导0有点麻烦; 即 10011 你不好获取 0011这个数字的信息: 即是不是到了从右往左数第二个0的时候,就该停下…
http://codeforces.com/problemset/problem/758/D 题意:给出一个进制数n,还有一个数k表示在n进制下的值,求将这个数转为十进制最小可以是多少. 思路:模拟着做,有点像two-pointer的做法.正着扫这个字符串,如果找到一个符合题意的即比n小的数,那么这个数是合法的,可以加上,接下来下标就移动到扫到的位置-1.主要难点我觉得在于0的时候的情况,如果当前的下标指的位置是0的话,无疑这个0是对当前的数是没有贡献的,那么这一位0应该不算上去,指针应该向后找…
D. Ability To Convert time limit per test 1 second Cmemory limit per test 256 megabytes input standard input output standard output Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English le…
D - Ability To Convert 题目大意:给你一个数字 n 接下来再输入一个数字 w(<10^60),表示w这个数字是 n 进制的, 并且超过十进制也用数字表示,这样就有多种组合了,问你所有组合中(划分方案中)原来的 数字十进制最小是多少. 思路:网上说可以用贪心做,但是我感觉那个贪心比较奇怪,不是很理解,还是用dp靠谱. 用dp[ i ] 表示从到第 i 个数字为止,最小十进制数最小是多少. 初始状态 dp[ 0 ]=0.然后我们枚举起点向后更新. #include<bits/…
D. Ability To Convert time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English let…
http://codeforces.com/contest/758/problem/D D. Ability To Convert time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexander is learning how to convert numbers from the decimal system to any…
D. Ability To Convert time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English let…
题目链接:http://codeforces.com/problemset/problem/758/D 题意:一个n进制下的数k,其中k不会用字母,如果有A就用10代替了.求k这个数对应的,在10进制下最小的数. 分析: 本质上是把数字分成若干段使得每一段 <n 且没有前导 0 dp[i] 表示前 i 个字符划分好之后能得到的最小数. 状态枚举下一段怎么切. #include<cstdio> #include<cstring> #include<cstdlib>…
在N进制下给你一个数,要你转换成最小的十进制数; 状态转移方程:从前向后 dp[j]表示j位前数列的最小十进制数 dp[j]=min(dp[j],dp[i]*n+x) 程序: #include <stdio.h> #include <iostream> #include <algorithm> #include <math.h> using namespace std; char s[1000]; long long dp[1000]; int main()…
Unfair Poll 题意:一共有n排同学每排同学有m个人, 老师问问题有一个顺序, 先从第一排开始问,问完第一排的所有同学之后,再问第2排的,对于所有排的访问顺序为 1,2,3……n-1,n,n-1,n-2,……,2,1,2,然后每次访问到新的一排先要问完这一排的所有人才会往下一(目标)排走. 题解:先声明我们开一个数组来记录这一排被询问的总次数,先将k  /= m, 这个代表的是完全访问的次数,即一整排m位同学都问完有几次,如果  完全访问的次数< n, 我们就将前几排全访问次数的人都加上…
link 题意:给定进制数n及一串数字,问在此进制下这串数能看成最小的数(10进制)是多少(如HEX下 1|13|11 = 475) 思路:此题要仔细思考细节.首先要想使数最小那么必定有个想法是使低位的数尽可能大即位数尽可能多,个人想法是从最后一位开始向前遍历,如果位数没有超过n的位数且数值比n小,那么该数可以加入上一个分隔.如果不符合条件,则说明该数应该作为新的分隔. 那么在此就有一个问题了,如果一个分隔是以0开始的(如:10000)显然我们也将0认定为是分隔的一部分,但如果有前导0了(11进…
从后往前贪心就好了.各种各样0的情况太BT了.. (各种爆long long,f**k) #include<bits/stdc++.h> #define LL long long #define N 100005 #define lowbit(x) x&(-x) using namespace std; inline int ra() { ,f=; char ch=getchar(); ; ch=getchar();} +ch-'; ch=getchar();} return x*f;…
D. Ability To Convert time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English let…
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D. Ability To Convert time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English let…
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D. Ability To Convert time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English let…
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