hdu 5666 (大数乘法) Segment

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hdu 5666 (大数乘法) Segment

题目:这里题意:在线段x+y=q与坐标轴围成的三角形中,求有多少个坐标为整数的点,答案模上p. 很容易就想到最后答案就是((q-1)*(q-2))/2然后模上p就是了,但是这个数字比较大,相乘会爆long long,于是用二进制的乘法,类似于快速幂,另外注意这个除以2得在之前就处理了,因为最后答案一定要是个准确的整数. #include<cstdio> #include<cstring> #include<iostream> #include<algorith…

HDU 1402 大数乘法 FFT、NTT

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.…

hdu-5666 Segment(俄罗斯乘法or大数乘法取模)

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…

HDOJ-1042 N!（大数乘法）

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…

51nod 1027大数乘法

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++;…

[POJ] #1001# Exponentiation : 大数乘法

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…

用分治法实现大数乘法，加法，减法（java实现）

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…