标签：
题目：POJ 2398
|
Bull Math
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 Source |
嘴笨，直接上代码。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;
const int maxn = 100;
void reverse(char a[])
{
int len = strlen(a);
for(int i = 0 ; i < len / 2; i++)
{
int temp = a[i];
a[i] = a[len - i - 1];
a[len - i -1] = temp;
}
}
int main()
{
char a[maxn],b[maxn];
int t[100] = {0};
//printf("Please enter 2 numbers: ");
scanf("%s%s",a,b);
reverse(a);
reverse(b);
if(strcmp(a,"0")==0||strcmp(b,"0")==0)
cout<<"0"<<endl;
else
{
int i,j;
for(i = 0; i <strlen(b); i++)
{
int cnt = 0;
for(j = 0; j < strlen(a); j++)
{
int temp = (b[i] - ‘0‘) * (a[j] - ‘0‘);
int tt= t[i+j] + temp + cnt;
t[j+i] = tt % 10;
cnt = tt / 10;
}
while(cnt != 0)
{
t[j+i] = cnt % 10;
cnt = cnt / 10;
j++;
}
}
for(int k = i + j - 2; k >= 0; k--)
{
cout<<t[k];
}
}
return 0;
}
标签：
原文地址：http://www.cnblogs.com/yqbeyond/p/4390998.html
