POJ 2389 Bull Math(大数乘大数)

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 two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros).

FJ asks that you do this yourself; don’t use a special library function for the multiplication.
Input

Lines 1…2: Each line contains a single decimal number.
Output
Line 1: The exact product of the two input lines

Sample Input

11111111111111
1111111111
Sample Output

12345679011110987654321
代码是模拟手算过程

#include<stdio.h>
#include<string.h>
int  read(int *am)
{int i=0;char st=getchar();while(st!='\n'){am[++i]=st-'0';st=getchar();}return i;
}
int main(void)
{int n1[50]={0};int n2[50]={0};int ans[100]={0};int carry=0;int size1=0,size2=0;size1=read(n1);size2=read(n2);int size=0;int k=0,tep;for(int j=size2;j>=1;j--,++k){size=k;//模拟加+for(int i=size1;i>=1;i--){tep=carry+ans[size]+n1[i]*n2[j];//顺便把上一次结果加进来，模拟运算ans[size++]=tep%10;carry=tep/10;}while(carry){ans[size++]=carry%10;carry/=10;}}for(int i=size-1;i>=0;i--){printf("%d",ans[i]);}
}

POJ 2389 Bull Math(大数乘大数)相关推荐

POJ 2389 Bull Math(水~Java -大数相乘）
题目链接:http://poj.org/problem?id=2389 题目大意: 大数相乘. 解题思路: java BigInteger类解决 o.0 AC Code: 1 import java. ...
POJ 2389 Bull Math（FFT）
Description 给出两个数A和B,求A*B Input 两个数字串,串长均不超过40 Output 输出两个串所表示数字的乘积 Sample Input 11111111111111 1111 ...
高精度乘法（正负数皆可(Bull Math)POJ）
描述高精度乘法输入:两行,每行表示一个非负整数(不超过10000位) 输出:两数的乘积. 样例1 样例输入1 99 101 样例输出1 9999 #include<iostream> ...
【bzoj 1754】【POJ - 2389 】Bull Math （高精度运算）
题干: Bulls are so much better at math than the cows. They can multiply huge integers together and get ...
NUC1372 Bull Math【大数】
Bull Math 时间限制: 1000ms 内存限制: 65535KB 通过次数: 1总提交次数: 1 问题描述 Bulls are so much better at math than the ...
**【POJ - 2389】 Bull Math （高精度乘法）
题干: Bulls are so much better at math than the cows. They can multiply huge integers together and get ...
poj2389 Bull Math （高精度之A*B）
题目来源:poj2389 Bull Math Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13350 Accepted ...
易语言大数计算大数支持库的使用实例
部分数据过大需要使用到大数计算,网上暂时没看到简单的例子实例,所以记录下.希望能帮上需要的人. 以下代码直接复制到易语言即可使用.支持库里需要勾选上数值计算支持库 EC模块下载地址:易语言大数 ...
【POJ - 1001 】Exponentiation （Java大数，高精度）
题干: Problems involving the computation of exact values of very large magnitude and precision are com ...

POJ 2389 Bull Math(大数乘大数)

POJ 2389 Bull Math(大数乘大数)相关推荐

最新文章

热门文章