Problem F: Exponentiation大数求幂】的更多相关文章

DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems. This problem requires that you write a…
Problem F: ExponentiationTime Limit: 1 Sec Memory Limit: 128 MBSubmit: 4 Solved: 2[Submit][Status][Web Board] [Edit] [TestData]DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example…
Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.     This problem requires that you wr…
Home Web Board ProblemSet Standing Status Statistics   Problem F: 求平均年龄 Problem F: 求平均年龄 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 720  Solved: 394[Submit][Status][Web Board] Description 定义一个Persons类,用于保存若干个人的姓名(string类型)和年龄(int类型),定义其方法 void ad…
接触ACM没几天,向各路大神求教,听说ACM主要是研究算法,所以便开始了苦逼的算法学习之路.话不多说,RT所示,学习快速求幂. 在头文件<math.h>或是<cmath>中,double pow( double x, double y );函数是用来快速求x^y,于是便从pow函数来说起,以下大体上是pow的函数代码: int pow(int x, int n) { int num = 1; while (n != 0){ num = num *x; n = n -1; } ret…
Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5930    Accepted Submission(s): 4146 Problem Description As we know, Big Number is always troublesome. But it's really important in our…
hdu-4549 求幂大法.矩阵快速幂.快速幂 题目 M斐波那契数列 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 6217 Accepted Submission(s): 1902 Problem Description M斐波那契数列F[n]是一种整数数列,它的定义如下: F[0] = a F[1] = b F[n] = F[n-1] *…
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高精度求幂 public static char[] exponentiation(string a,int r) { ]; string b = ""; string c = a; ; i < r-; i++) { aa = acm.Quadrature(c, a); b = ""; foreach (var item in aa) { b += item; } c = b; } return aa; }…
题目地址:http://ac.jobdu.com/problem.php?pid=1085 题目描述: N<k时,root(N,k) = N,否则,root(N,k) = root(N',k).N'为N的k进制表示的各位数字之和.输入x,y,k,输出root(x^y,k)的值 (这里^为乘方,不是异或),2=<k<=16,0<x,y<2000000000,有一半的测试点里 x^y 会溢出int的范围(>=2000000000) 输入: 每组测试数据包括一行,x(0<…