题意 You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numb…

把输入数字每次从9-2除,能整除则记录该数字,最后从小到大输出. 应该算是水题,不过窝第一次写高精度除法,虽然1A,不过中间改了好多次. /****************************************** Problem: 2325 User: Memory: 684K Time: 110MS Language: G++ Result: Accepted ******************************************/ #include <iostrea…

(-￣▽￣)-* 这道题涉及高精度除法,模板如下: ]; ];//存储进行高精度除法的数据 bool bignum_div(int x) { ,num=; ;s[i];i++) { num=num*+s[i]-'; division[tot++]=num/x+'; num%=x; } division[tot]='\0';//利于进行strcpy() ) //有适合的除数 { ; ') i++; strcpy(s,division+i);//比如49->07,那么下一轮s就变成7,多余的i个0都…

题意:给你一个数n,让你找m个非负整数,使得它们的和为n,并且按位或起来以后的值最小化.输出这个值. 从高位到低位枚举最终结果,假设当前是第i位,如果m*(2^i-1)<n的话,那么说明这一位如果填零,剩下的位不论怎么填,都绝对凑不出n来,所以这一位必须填1.如果m*(2^i-1)>=n,这一位就填1,然后把n变为max(n mod 2^i , n-m*2^i),重复这个过程即可. 要用高精度. import java.util.*; import java.io.*; import java…

题意:给两个整数,求这两个数的反向数的和的反向数,和的末尾若为0,反向后则舍去即可.即若1200,反向数为21.题目给出的数据的末尾不会出现0,但是他们的和的末尾可能会出现0. #include <iostream> #include <string.h> #include <stdio.h> #include <string> #include <string.h> using namespace std; int n1,n2;//n1表示a的…

题目链接 题意 :与1009一样,不过这个题的数据范围变大. 思路:因为数据范围变大,所以要用大数模拟,用java也行,大数模拟也没什么不过变成二维再做就行了呗.当然也可以先把所有的都进行打表,不过要用三维了就 . //URAL 1012 #include <stdio.h> #include <string.h> #include <iostream> using namespace std ; ]; ][] ; int main() { int n, k ; sca…

Censored! Time Limit: 5000MS   Memory Limit: 10000K Total Submissions: 6956   Accepted: 1887 Description The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters w…

树状数组维护DP + 高精度 Description These days, Sempr is crazed on one problem named Crazy Thair. Given N (1 ≤ N ≤ 50000) numbers, which are no more than 109, Crazy Thair is a group of 5 numbers {i, j, k, l, m} satisfying: 1 ≤ i < j < k < l < m ≤ N Ai…

How many Fibs? Description Recall the definition of the Fibonacci numbers: f1 := 1 f2 := 2 fn := f n-1 + f n-2 (n>=3) Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a,b]. Input The input contains several test cases.…

题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=169 解题报告: P(n)定义为n的所有位数的乘积,例如P(1243)=1*2*3*4=24,然后如果P(n)!=0且n mod P(n) = 0,则称n为good number.如果n和n+1都为good numbers,则称n为perfect number.然后给出位数k(1<=k<=1000000),找出位数为k的perfect number.题目给出的k范围很大,我每次看…