IYukinaI @ 2024-10-17 17:29:44

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long a,b;
    cin>>a>>b;
    cout<<a*b;
    return 0;
}

by I_Love_DS @ 2024-10-17 17:31:43

6

需要高精度

by jqQt0220 @ 2024-10-17 17:32:29

每个非负整数不超过 10^{2000}。

by HeiCat0725 @ 2024-10-17 17:43:21

高精度或者如果你没学高精度可以使用py

by YDMaYi @ 2024-10-17 17:44:13

@IYukinaI

人生苦短，我用python

by HeiCat0725 @ 2024-10-17 17:46:00

@IYukinaI

a=int(input())
b=int(input())
print(a*b)

这是py代码，但是还是建议学习一下高精度。

by CEFqwq @ 2024-10-17 17:49:24

@IYukinaI 写个 FFT 不好吗

by IYukinaI @ 2024-10-18 12:48:37

蟹蟹٩('ω')و

by hzy6622 @ 2024-10-25 20:40:18

我也是这么写的```c

include <bits/stdc++.h>

using namespace std; int main(){ long long a,b; cin>>a>>b; cout<<a*b; return 0; }