<模板> Hdu 1402 A * B Problem Plus 大数乘法
发布时间：2020-12-14 03:05:19 所属栏目：大数据 来源：网络整理
导读：A * B Problem Plus Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12315????Accepted Submission(s): 2157 Problem Description Calculate A * B. ? Input Each line will contain two integ
A * B Problem PlusTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12315????Accepted Submission(s): 2157
Problem Description
Calculate A * B.
?
Input
Each line will contain two integers A and B. Process to end of file.
Note: the length of each integer will not exceed 50000.
?
Output
For each case,output A * B in one line.
?
Sample Input
?
Sample Output
?
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#define LL __int64
#define M 10000000000
using namespace std;
LL a[5100],b[5100],c[110000];
char s1[500010],s2[500010];
void MUL()
{
int t,i,j;
memset(c,sizeof(c));
for (i=a[0];i>0;i--)
{
t=a[0]-i+1;
for (j=b[0];j>0;j--)
{
c[t+b[0]-j+1]+=a[i]*b[j] / M;
c[t+b[0]-j]+=a[i]*b[j]%M;
}
if (c[t+1])
c[0]=t+1;
else c[0]=t;
}
}
int main()
{
while (gets(s1))
{
gets(s2);
strrev(s1);
strrev(s2);
int l1=strlen(s1)-1;
int l2=strlen(s2)-1;
memset(a,sizeof(0));
memset(b,sizeof(0));
while(l1>=0)
{
LL k=0,s=0;
while (k<10 && l1>=0)
{
s=s*10+s1[l1]-'0';
k++;
l1--;
}
a[0]++;
a[a[0]]=s;
}
while(l2>=0)
{
int k=0,s=0;
while (k<10 && l2>=0)
{
s=s*10+s2[l2]-'0';
k++;
l2--;
}
b[0]++;
b[b[0]]=s;
}
MUL();
printf("%I64d",c[c[0]]);
for (int i=c[0]-1;i>1;i--)
printf("%010I64d",c[i]);
printf("%I64dn",c[1]);
//cout<<endl;
}
return 0;
}
（编辑：李大同） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |