<模板> Hdu 1402 A * B Problem Plus 大数乘法-优快云博客

本文链接：https://blog.youkuaiyun.com/modiz/article/details/38499939

本文介绍了一种处理大整数相乘问题的算法实现，针对长度不超过50,000位的大整数，通过特殊的输入处理和乘法运算策略来避免普通整数类型无法直接处理的问题。

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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 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

2
2000

虽然还是没有AC。超时了。未解之谜，仔细再看看。

#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,0,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,0,sizeof(0));
		memset(b,0,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("%I64d\n",c[1]);
		//cout<<endl;
	}
	return 0;
}