这道题虽然数据很大，但是可以用数学的乘法运算来解决
Description
Calculate a*b
Input
Input contains multiple test cases.
Each test case contains two number a and b (0<=a,b<=10^1000),
Output
Output a*b
Sample Input
12345678987654321 98765432123456789
Sample Output
1219326320073159566072245112635269
Hint
练习模拟大数乘法
#include <cstdio>
#include <iostream>
#include <cmath>
#include <
string>
#include <cstring>
#include <algorithm>
using namespace std;
string a, b;
int x[
1000+
8], y[
1000+
8], sum[
1000000+
8], t;
int main()
{
while(cin >> a >>
b)
{
int lena =
a.size();
int lenb =
b.size();
int lens =
0;
memset(sum, 0,
sizeof(sum));
for(
int i =
0; i<lena; i++
)
{
x[lena-i] = a[i]-
48;
//-48等于-'0',并且把这个字符串倒序输入数列X中
}
for(
int i =
0; i<lenb; i++
)
{
y[lenb-i] = b[i]-
48;
}
for(
int i =
1; i<=lena; i++)
//像数学乘法一样，把所有的数都乘起来
{
t =
0;
//记录要进的位数
for(
int j =
1; j <= lenb; j++
)
{
sum[i+j-
1] += x[i]*y[j]+
t;
t = sum[i+j-
1]/
10;
sum[i+j-
1] %=
10;
}
sum[i+lenb] =
t;
}
lens = lena+
lenb;
while(sum[lens] ==
0 &&lens>
1) lens--
;
for(
int i = lens; i >=
1; i--)printf(
"%d", sum[i]);
printf("\n");
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/10371456.html
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