http://acm.hdu.edu.cn/showproblem.php?pid=1402 快速傅里叶变换优化的高精度乘法. https://blog.csdn.net/ggn_2015/article/details/68922404 这个写的很详细了. #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<iostream> #incl…

/* hdu 1402 A * B Problem Plus FFT 这是我的第二道FFT的题 第一题是完全照着别人的代码敲出来的,也不明白是什么意思 这个代码是在前一题的基础上改的 做完这个题,我才有点儿感觉,原来FFT在这里就是加速大整数乘法而已 像前一题,也是一个大整数乘法,然后去掉一些非法的情况 */ #pragma warning(disable : 4786) #pragma comment(linker, "/STACK:102400000,102400000") #in…

题意:大数乘法 思路:FFT模板 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81…

因为刚学fft,想拿这题练练手,结果WA了个爽= =. 总结几点犯的错误: 1.要注意处理前导零的问题. 2.一定要注意数组大小的问题.(前一个fft的题因为没用到b数组,所以b就没管,这里使用了b数组,结果忘记给其大小乘以4倍了) 代码如下: #include<bits/stdc++.h> using namespace std; ; ; typedef long long ll; struct Complex { double x,y; Complex(,) :x(_x),y(_y) {}…

A * B Problem Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9413    Accepted Submission(s): 1468 Problem Description Calculate A * B.   Input Each line will contain two integers A and B.…

A * B Problem Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26874    Accepted Submission(s): 7105 Problem Description Calculate A * B.   Input Each line will contain two integers A and B.…

题意:计算A*B,A,B均为长度小于50000的整数. 这是FFT在大整数相乘中的一个应用,我本来想用NTT做的,但NTT由于取模很可能取炸,所以base必须设得很小,而且效率也比不上FFT. A和B的存储均用long long,在计算乘积的时候转化成double,计算完成后再转回来即可. 测得base在精度允许范围内最多能开到10000. 把平方和快速幂的函数也写上了,可以当模板用~ #include<bits/stdc++.h> using namespace std; typedef l…

http://acm.hdu.edu.cn/showproblem.php?pid=1402 题意: 求$a*b$ 但是$a$和$b$的范围可以达到 $1e50000$ 题解: 显然...用字符串模拟的大数或者压位的大数是无法胜任这种计算的.... 然后,2个大整数相乘,可以理解为卷积,所以就用快速傅里叶变换(FFT)来加速他 模板题 简单总结一下对FFT的认知: FFT用于算卷积,卷积可以理解为两个多项式相乘显然复杂度是$O(n^2)$的 但是fft可以优化为$O(nlogn)$如何优化,考虑…

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.   题目大意:求A * B. 思路:快速傅里叶变换的模板题,…

A * B Problem Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16932    Accepted Submission(s): 3558 Problem Description Calculate A * B.   Input Each line will contain two integers A and B…

A * B Problem Plus 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…

猜公式: ans=n/m^(n-1) #include<stdio.h> #include<string.h> struct BigNum { ]; int len; }; int gcd(int a,int b) { ) return a; return gcd(b,a%b); } BigNum mul(BigNum &a,int b) { BigNum c; int i,len; len=a.len; memset(c.num,,sizeof(c.num)); ) {…

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 唉,模板题,膜的邝…

FFT模板题,求A*B. 用次FFT模板需要注意的是,N应为2的幂次,不然二进制平摊反转置换会出现死循环. 取出结果值时注意精度,要加上eps才能A. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; const double pi = acos(-1.0); const i…

$A * B$ FFT模板题,找到了一个看起来很清爽的模板 /** @Date : 2017-09-19 22:12:08 * @FileName: HDU 1402 FFT 大整数乘法.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <bits/stdc++.h> #define LL…

Train Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10372    Accepted Submission(s): 5543 Problem Description As we all know the Train Problem I, the boss of the Ignatius Train Sta…

数字的反转: 就是将数字倒着存下来而已.(*^__^*) 嘻嘻…… 大致思路:将数字一位一位取出来,存在一个数组里面,然后再将其变成数字,输出. 详见代码. while (a) //将每位数字取出来,取完为止 { num1[i]=a%; //将每一个各位取出存在数组里面,实现了将数字反转 i++; //数组的变化 a/=; } 趁热打铁 例题:hdu 4554 叛逆的小明 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4554 叛逆的小明 Time…

A * B Problem Plus HDU - 1402 (FFT) Calculate A * B.  InputEach line will contain two integers A and B. Process to end of file. Note: the length of each integer will not exceed 50000. OutputFor each case, output A * B in one line. Sample Input 1 2 10…

A * B Problem Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9413    Accepted Submission(s): 1468 Problem Description Calculate A * B.   Input Each line will contain two integers A and B.…

大数乘法问题一般可以通过将大数转换为数组来解决. 解题思路 第1步 第2步 第3步 第4步 样例输入1 56 744 样例输出1 800 样例输入2 -10 678 样例输出2 -6780 样例输入3 1234567890 45678901234 样例输出3 56393704713977776260 代码实现 #include<stdio.h> #include<string.h> #define MAX 1000 // 大数乘法 void Multiply(char* tempA…

http://acm.hdu.edu.cn/showproblem.php?pid=1402 fft做O(nlog(n))大数乘法,kuangbin的模板 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> using namespace std; const double PI = acos(-1.0)…

p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-size: 10.5000pt } h1 { margin-top: 5.0000pt; margin-bottom: 5.0000pt; text-align: center; font-family: 宋体; color: rgb(26,92,200); font-weight: bold; fo…

一. 题目 Exponentiation Time Limit: 500MS   Memory Limit: 10000K Total Submissions: 156373   Accepted: 38086 Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of…

http://acm.hdu.edu.cn/showproblem.php?pid=1042 题意清晰..简单明了开门见山的大数乘法.. 10000的阶乘有35000多位 数组有36000够了 # include <stdio.h> # include <string.h> # define MAX 36000 int BigNum[MAX], NowLen; void Multi(int number) { int Temp[MAX]={0}, Tlen = 0, t;//Tem…

Product Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description The problem is to multiply two integers X, Y. (0<=X,Y<10250) Input The input will consist of a set of pairs of lines. Each line in pair cont…

题目链接: Segment Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description       Silen August does not like to talk with others.She like to find some interesting problems. Today she finds an interesting pro…

Problem Description A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a ~{!0~}cycle~{!1~} of the digits of the original number. That is, if you consider the number after the last digit to ~{!0~…

题目链接:51nod 1027大数乘法 直接模板了. #include<cstdio> #include<cstring> using namespace std; ; ; ; int alen, blen; int ans_len; char a1[N], b1[N]; ], b[]; ]; ], int &len){ memset(c, , sizeof(c)); int L = strlen(s); len = L / DLEN; if(L%DLEN) len++;…

大数乘法即多项式乘法问题,求A(x)与B(x)的乘积C(x),朴素解法的复杂度O(n^2),基本思想是把多项式A(x)与B(x)写成 A(x)=a*x^m+b B(x)=c*x^m+d 其中a,b,c,d为x的多项式. 则A(x)*B(x)=(ac)*x^2m+(ad+bc)*x^m+bd 由ad+bc=(a+b)(c+d)-ac-bd 原来的4次乘法和1次加法由3次乘法和2次减法代替,减少了一次乘法操作. 用同样的方法应用到abcd的乘法上. (以上内容摘自互联网) 以下为用java实现的代码…

HDU 4291 A Short problem(2012 ACM/ICPC Asia Regional Chengdu Online) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4291 Description 给一个式子求结果.类似Fibonacci的公式g(n)=3*g(n-1)+g[n-2]. Input 给你n(1<=n<=1e18) Output 求g(g(g(n))) Sample Input 样例第一个就是0什么鬼,虽然没影响.…