45 lines
1.0 KiB
C++
45 lines
1.0 KiB
C++
|
#include <bits/stdc++.h>
|
|||
|
|
|
|||
|
|
using namespace std;
|
|||
|
|
|
|||
|
|
/**
|
|||
|
|
* 功能：高精度乘高精度模板
|
|||
|
|
* @param A
|
|||
|
|
* @param b
|
|||
|
|
* @return
|
|||
|
|
*/
|
|||
|
|
vector<int> mul(vector<int> &A, vector<int> &B) {
|
|||
|
|
//初始化大小
|
|||
|
|
vector<int> C(A.size() + B.size());
|
|||
|
|
//先放里再说
|
|||
|
|
for (int i = 0; i < A.size(); i++)
|
|||
|
|
for (int j = 0; j < B.size(); j++)
|
|||
|
|
C[i + j] += A[i] * B[j];
|
|||
|
|
|
|||
|
|
//处理余数
|
|||
|
|
for (int i = 0, t = 0; i < C.size(); i++) {
|
|||
|
|
t += C[i];
|
|||
|
|
if (i >= C.size()) C.push_back(t % 10);
|
|||
|
|
else C[i] = t % 10;
|
|||
|
|
t /= 10;
|
|||
|
|
}
|
|||
|
|
//去掉前导0
|
|||
|
|
while (C.size() > 1 && C.back() == 0) C.pop_back();
|
|||
|
|
return C;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
int main() {
|
|||
|
|
string a, b;
|
|||
|
|
cin >> a >> b;
|
|||
|
|
//准备动作
|
|||
|
|
vector<int> A, B;
|
|||
|
|
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
|
|||
|
|
for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');
|
|||
|
|
//计算
|
|||
|
|
vector<int> C = mul(A, B);
|
|||
|
|
//倒序输出
|
|||
|
|
for (int i = C.size() - 1; i >= 0; i--) cout << C[i];
|
|||
|
|
return 0;
|
|||
|
|
}
|
|||
|
|
|