跪求大佬
by The_fisher @ 2024-01-29 11:46:01
@[zhaoyuhang1231](/user/1011497) 我问你 $(10^{101}-1)\times(10^{101}-1)$ 有几位
by Special_Tony @ 2024-01-29 11:57:52
@[zhaoyuhang1231](/user/1011497) 另外你不看数据范围吗，2000的范围你只开100，不RE才怪
by Special_Tony @ 2024-01-29 11:58:29
@[zhaoyuhang1231](/user/1011497)
代码帮你改好了
```cpp
#include<bits/stdc++.h>
using namespace std;
char a1[50001],b1[50001];
int a[50001],b[50001],c[50001],lena,lenb,lenc,i,j;//数组开大一点就不会RE了
//放在外面就不用初始化啦~
int main()
{
scanf("%s%s",a1,b1);
lena=strlen(a1);lenb=strlen(b1);
for (i=1;i<=lena;i++) a[i]=a1[lena-i]-'0';
for (i=1;i<=lenb;i++) b[i]=b1[lenb-i]-'0';
//是这样写哦
for (i=1;i<=lena;i++)
{
for (j=1;j<=lenb;j++)
{
c[i+j-1]+=a[i]*b[j]; //要注意是+=
}
}
for(i=1;i<lena+lenb;i++)
if(c[i]>9)
{
c[i+1]+=c[i]/10;
c[i]%=10;
}
//处理进位在外面会方便一点
lenc=lena+lenb;
while (c[lenc]==0&&lenc>1)
lenc--;
for (i=lenc;i>=1;i--)
cout<<c[i];
cout<<endl;
return 0;
}
```
by 2023tanglehao @ 2024-01-29 12:05:41
谢谢大佬们
关注瞻仰
by The_fisher @ 2024-01-29 13:54:36