http://www.bnuoj.com/bnuoj/problem_show.php?pid=29375 [题意]:可以对两字符串进行如下操作: 1.可以无损耗交换相邻两个字符(可以理解成交换任意字符) 2.可以改变一个字符 x->y ,花费为 x-y 的绝对值 求花费最少,将两字符串变成一样 [题解]: 排序字符串,然后对应相减 [code]: #include <iostream> #include <stdio.h> #include <math.h> #…

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero.…

Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 这道题让我们求两个字符串数字的相乘,输入的两个数和返回的数都是以字符串格式储存的,这样做的原因可能是这样可以计算超大数相乘,可以不受int或long的数值范围的约束,那么我们该如何来计算…

题目链接: 题目 D. Alyona and Strings time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output 问题描述 After returned from forest, Alyona started reading a book. She noticed strings s and t, lengths of which are…

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123&…

转载:43. Multiply Strings 题目描述 就是两个数相乘,输出结果,只不过数字很大很大,都是用 String 存储的.也就是传说中的大数相乘. 解法一 我们就模仿我们在纸上做乘法的过程写出一个算法. 个位乘个位,得出一个数,然后个位乘十位,全部乘完以后,就再用十位乘以各个位.然后百位乘以各个位,最后将每次得出的数相加.十位的结果要补 1 个 0 ,百位的结果要补两个 0 .相加的话我们可以直接用之前的大数相加.直接看代码吧. public String multiply(Stri…

Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 给出两个字符串,返回对应数字想乘后的字符串,由于这个字符串可能很大,所以不能采用一般的乘法,这里用的方法是模拟手工的乘法运算,算法 本身很简单,就是当时写的时候有些很小的细节搞错了,找了…

Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 题意:计算字符串对应数字的乘积.参考JustDoIT的博客. 思路:先用一维向量存放计算的中间值,中间值存放对应位置乘积的和(暂不考虑进位的情况):然后针对进位的情况,进行计算:注意要跳…

Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 要求:字符串表示的数字可能无穷大,并且非负. class Solution { private: vector<string> tempStrs; public: string add…

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero.…