算法爬坑之线性表大数乘法
WAI0`%JDC1P({JBR$DE{F}O.jpginclude<iostream>
include<memory.h>
include<string.h>
using namespace std;
int* multi(int* num1, int size1, int* num2, int size2){
int size = size1 + size2;
int* ret = new int[size];
int i = 0;
memset(ret, 0, sizeof(int) * size);
for(i = 0; i<size2; ++i){
int k = i;
for(int j = 0; j < size1; ++j){
ret[k++] += num2[i] * num1[j];
}
}
for (i = 0; i < size; ++i){
if(ret[i] >= 10){
ret[i+1] += ret[i]/10;
ret[i] %= 10;
}
}
return ret;
}
int main(int argc, char** argv) {
char digit1[100];
char digit2[100];
cin >> digit1 >> digit2;
int size1 = strlen(digit1);
int size2 = strlen(digit2);
int* num1 = new int[size1];
int* num2 = new int[size2];
int* currentDigit1 = new int[size1];
int* currentDigit2 = new int[size2];
for(int j = 0, jj = size1 - 1; j < size1; j++, jj--){
num1[jj] = digit1[j] - '0';
}
for(int k = 0, kk = size2 - 1; k < size2; k++, kk--){
num2[kk] = digit2[k] - '0';
}
int* ret = multi(num1, size1, num2, size2);
cout<<"求得的乘积为:"<<endl;
for(int i = size1+size2-2; i >= 0 ; i--){
cout << ret[i];
}
delete[] ret;
return 0;
}
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