题目连接:1220 NUMBER BASE CONVERSION 题目大意:给出两个进制oldBase 和newBase, 以及以oldBase进制存在的数.要求将这个oldBase进制的数转换成newBase进制的数. 解题思路:短除法,只不过时直接利用了高精度的除法运算, 并且将以前默认的*10换成的*oldBase. #include <stdio.h> #include <string.h> const int N = 1005; int newBase, oldBase,…

NUMBER BASE CONVERSION Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5976   Accepted: 2738 Description Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits: { 0-9,A-Z,a-z } HINT: If…

题目链接 题意 : 给你一个a进制的数串s,让你转化成b进制的输出. A = 10, B = 11, ..., Z = 35, a = 36, b = 37, ..., z = 61,0到9还是原来的含义. 思路 : 这个题因为牵扯了英文字母所以比较复杂一点,先将所有出现的英文字母转化成他们所代表的数,然后进行进制转换. //POJ 1220 #include <stdio.h> #include <string.h> #include <iostream> #incl…

package com.njupt.acm; import java.math.BigInteger; import java.util.Scanner; public class POJ_1220_1 { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int t = scanner.nextInt(); while(t > 0){ BigInteger ba1 = scann…

JavaScript Number对象 是经过封装的能处理数字值的对象 由Number()构造器创建 只有一种数字类型 可以使用也可以不使用小数点书写数字 JavaScript原始值与对象: 在JavaScript中,除了原始值都是对象 原始值通常包括数字.字符串.布尔值(如果不是用new Object()形式定义的话) 原始值不包含方法和属性 JavaScript数值:始终是64位的浮点数 与其他类型的语言不同,JavaScript始终以64位双精度浮点数来存储数值,0-51存储值,52-62…

Basic remains Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5221   Accepted: 2203 Description Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p mod m is defined as…

Description Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits: { 0-9,A-Z,a-z } HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next, when…

Judge Info Memory Limit: 32768KB Case Time Limit: 1000MS Time Limit: 1000MS Judger: Number Only Judger Description 现在假设: 2进制对应的基数是0,1; 3进制对应的基数是0,1,2; …… 10进制对应的基数是0,1,2,3,4,5,6,7,8,9. …… n进制对应的基数是0,1,2,3,……,n-1; 你的任务是实现进制之间的转换. Input 第一行,有t(t<=1000)…

$Poj$   $AcWing$ $Description$ $Sol$ 进制转化+高精度除法 $over$ $Code$ #include<bits/stdc++.h> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,size…

题目链接:http://poj.org/problem?id=2305 ime Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5326   Accepted: 2267 Description Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p…