https://codeforces.com/contest/1181/problem/B 从中间拆开然后用大数搞一波. 当时没想清楚奇偶是怎么弄,其实都可以,奇数长度字符串的中心就在len/2,偶数长度字符串的中心恰好是len/2和len/2-1. 但是要是作为末尾指针的位置来说的话 奇数长度字符串:把中心分给后半串,那么分割点是len/2.把中心分给前半串,那么分割点是len/2+1. 偶数长度字符串:分割点是len/2. 注意子串的取法,其实最方便的还是传头尾指针进去. #include<…

题目描述: time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Dima worked all day and wrote down on a long paper strip his favorite number n consisting of l digits. Unfortunately, the strip turned…

大数据加法给一个数num和最大迭代数k每次num=num+num的倒序,判断此时的num是否是回文数字,是则输出此时的数字和迭代次数如果k次结束还没找到回文数字,输出此时的数字和k 如果num一开始是回文数字,那么直接输出num和0即可. #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <string> using na…

B. Split a Number time limit per test2 seconds memory limit per test512 megabytes inputstandard input outputstandard output Dima worked all day and wrote down on a long paper strip his favorite number n consisting of l digits. Unfortunately, the stri…

Split a Number time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Dima worked all day and wrote down on a long paper strip his favorite number nn consisting of ll digits. Unfortunately, the s…

收藏的一段关于java大数运算的代码: package study_02.number; import java.math.BigDecimal; import java.math.BigInteger; public class BigNumber { // 默认除法运算精度,即保留小数点多少位 private static final int DEFAULT_DIV_SCALE = 10; // 这个类不能实例化 private BigNumber() { } /** * 提供精确的加法运算…

Description Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in tw…

Computer Transformation Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4561 Accepted: 1738 Description A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simul…

偶然又遇到了一道大数题,据说python大数运算好屌,试了一发,果然方便-1 a = int( raw_input() ); //注意这里是按行读入的,即每行只读一个数 b = int( raw_input() ); print a+b; print a*b; print a/b; print a%b;…

用JAVA 实现算术表达式(1234324234324 + 8938459043545)/5 + 343434343432.59845 因为JAVA语言中的long 定义的变量值的最大数受到限制,例如123456789987654321这样的整数就不能存放在long类型的变量中,如果这样两个大数相加或相乘,产生的结果会更大.比如,JAVA语言中如果使用long l = 1000000000这样定义没错,但如果加上2000000000变成 1000000000+2000000000测试结果就为-1…