大数乘法问题一般可以通过将大数转换为数组来解决. 解题思路 第1步 第2步 第3步 第4步 样例输入1 56 744 样例输出1 800 样例输入2 -10 678 样例输出2 -6780 样例输入3 1234567890 45678901234 样例输出3 56393704713977776260 代码实现 #include<stdio.h> #include<string.h> #define MAX 1000 // 大数乘法 void Multiply(char* tempA…

Train Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10372    Accepted Submission(s): 5543 Problem Description As we all know the Train Problem I, the boss of the Ignatius Train Sta…

一. 题目 Exponentiation Time Limit: 500MS   Memory Limit: 10000K Total Submissions: 156373   Accepted: 38086 Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of…

Product Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description The problem is to multiply two integers X, Y. (0<=X,Y<10250) Input The input will consist of a set of pairs of lines. Each line in pair cont…

题目链接: Segment Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description       Silen August does not like to talk with others.She like to find some interesting problems. Today she finds an interesting pro…

Problem Description A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a ~{!0~}cycle~{!1~} of the digits of the original number. That is, if you consider the number after the last digit to ~{!0~…

题目链接:51nod 1027大数乘法 直接模板了. #include<cstdio> #include<cstring> using namespace std; ; ; ; int alen, blen; int ans_len; char a1[N], b1[N]; ], b[]; ]; ], int &len){ memset(c, , sizeof(c)); int L = strlen(s); len = L / DLEN; if(L%DLEN) len++;…

大数乘法即多项式乘法问题,求A(x)与B(x)的乘积C(x),朴素解法的复杂度O(n^2),基本思想是把多项式A(x)与B(x)写成 A(x)=a*x^m+b B(x)=c*x^m+d 其中a,b,c,d为x的多项式. 则A(x)*B(x)=(ac)*x^2m+(ad+bc)*x^m+bd 由ad+bc=(a+b)(c+d)-ac-bd 原来的4次乘法和1次加法由3次乘法和2次减法代替,减少了一次乘法操作. 用同样的方法应用到abcd的乘法上. (以上内容摘自互联网) 以下为用java实现的代码…

http://acm.hdu.edu.cn/showproblem.php?pid=1042 题意清晰..简单明了开门见山的大数乘法.. 10000的阶乘有35000多位 数组有36000够了 # include <stdio.h> # include <string.h> # define MAX 36000 int BigNum[MAX], NowLen; void Multi(int number) { int Temp[MAX]={0}, Tlen = 0, t;//Tem…

1027 大数乘法 基准时间限制:1 秒 空间限制:131072 KB 分值: 0 难度:基础题 给出2个大整数A,B,计算A*B的结果. Input 第1行:大数A 第2行:大数B (A,B的长度 <= 1000,A,B >= 0) Output 输出A * B Input示例 123456 234567 Output示例 28958703552 题目链接:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1027 分…