A * B Problem Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26459 Accepted Submission(s): 6927
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
1
2
1000
2
Sample Output
2
2000
Author
DOOM III
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题目大意:
求两个大整数的乘积, 两个大整数长度都不超过50000, 多组数据, 时限1s
分析:
考虑把整数视作是两个多项式, 每一位就是一项, 那么就相当于是两个最高次数不超过50000的多项式,乘积之后在x = 10出的值, 直接处理出其多项式然后用FFT计算即可
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
const double PI = acos(-1.0);
struct Complex
{
double real, image;
Complex(double _real, double _image)
{
real = _real;
image = _image;
}
Complex(){}
};
Complex operator + (const Complex &c1, const Complex &c2)
{
return Complex(c1.real + c2.real, c1.image + c2.image);
}
Complex operator - (const Complex &c1, const Complex &c2)
{
return Complex(c1.real - c2.real, c1.image - c2.image);
}
Complex operator * (const Complex &c1, const Complex &c2)
{
return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}
int rev(int id, int len)
{
int ret = 0;
for(int i = 0; (1 << i) < len; i++)
{
ret <<= 1;
if(id & (1 << i)) ret |= 1;
}
return ret;
}
Complex A[140000];
void FFT(Complex* a, int len, int DFT)//对a进行DFT或者逆DFT, 结果存在a当中
{
//Complex* A = new Complex[len]; 这么写会爆栈
for(int i = 0; i < len; i++)
A[rev(i, len)] = a[i];
for(int s = 1; (1 << s) <= len; s++)
{
int m = (1 << s);
Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));
for(int k = 0; k < len; k += m)
{
Complex w = Complex(1, 0);
for(int j = 0; j < (m >> 1); j++)
{
Complex t = w*A[k + j + (m >> 1)];
Complex u = A[k + j];
A[k + j] = u + t;
A[k + j + (m >> 1)] = u - t;
w = w*wm;
}
}
}
if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;
for(int i = 0; i < len; i++) a[i] = A[i];
return;
}
char numA[50010], numB[50010];//以每一位为系数, 那么多项式长度不超过50000
Complex a[140000], b[140000];
//对应的乘积的长度不会超过100000, 也就是不超过(1 << 17) = 131072
int ans[140000];
int main()
{
while(~scanf("%s", numA))
{
int lenA = strlen(numA);
int sa = 0;
while((1 << sa) < lenA) sa++;
scanf("%s", numB);
int lenB = strlen(numB);
int sb = 0;
while((1 << sb) < lenB) sb++;
//那么乘积多项式的次数不会超过(1 << (max(sa, sb) + 1))
int len = (1 << (max(sa, sb) + 1));
for(int i = 0; i < len; i++)
{
if(i < lenA) a[i] = Complex(numA[lenA - i - 1] - '0', 0);
else a[i] = Complex(0, 0);
if(i < lenB) b[i] = Complex(numB[lenB - i - 1] - '0', 0);
else b[i] = Complex(0, 0);
}
FFT(a, len, 1);
FFT(b, len, 1);//把A和B换成点值表达
for(int i = 0; i < len; i++)//做点值表达的成乘法
a[i] = a[i]*b[i];
FFT(a, len, -1);//逆DFT换回原来的系数, 虚部一定是0
for(int i = 0; i < len; i++)
ans[i] = (int)(a[i].real + 0.5);//取整误差的处理
for(int i = 0; i < len - 1; i++)//进位问题
{
ans[i + 1] += ans[i] / 10;
ans[i] %= 10;
}
bool flag = 0;
for(int i = len - 1; i >= 0; i--)//注意输出格式的调整即可
{
if(ans[i]) printf("%d", ans[i]), flag = 1;
else if(flag || i == 0) printf("0");
}
putchar('\n');
}
return 0;
}