Power of Cryptography  Background Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of these primes. Work in this area has resulted in the practical use of results from num…

题目链接: https://vjudge.net/problem/POJ-2109 题目大意: 有指数函数 k^n = p , 其中k.n.p均为整数且 1<=k<=10^9 , 1<=n<= 200 , 1<=p<10^101 给定 n 和 p ,求底数 k 思路: 一开始以为需要大数,没想到一个pow就行了,真是涨姿势 考虑到数值存储问题和精度问题,这题最直观的思路应该是使用 高精度算法 求解.    而事实上,这题也可用公式法求解,但需要一些技巧. 开方公式:k…

Power of Cryptography Time limit: 3.000 seconds http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=49 http://poj.org/problem?id=2109   Time Limit: 1000MS   Memory Limit: 30000K Total…

[POJ2109]Power of Cryptography 试题描述 Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory a…

题目地址:http://poj.org/problem?id=2109 /* 题意:k ^ n = p,求k 1. double + pow:因为double装得下p,k = pow (p, 1 / n); 基础知识: 类型 长度 (bit) 有效数字 绝对值范围 float 32 6~7 10^(-37) ~ 10^38 double 64 15~16 10^(-307) ~ 10^308 long double 128 18~19 10^(-4931) ~ 10 ^ 4932 2. 二分查找…

点击打开链接 Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16388   Accepted: 8285 Description Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these pr…

1.Link: http://poj.org/problem?id=2109 http://bailian.openjudge.cn/practice/2109/ 2.Content: Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18872   Accepted: 9520 Description Current work in cryptography involves (…

Power of Cryptography DescriptionCurrent work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and…

uva 10330 - Power Transmission 题目大意:最大流问题. 解题思路:増广路算法. #include <stdio.h> #include <string.h> #include <queue> using namespace std; #define min(a,b) (a)<(b)?(a):(b) const int N = 105; const int INF = 0x3f3f3f3f; int n, s[N], g[N][N],…

UVA 11149 - Power of Matrix 题目链接 题意:给定一个n*n的矩阵A和k,求∑kiAi 思路:利用倍增去搞.∑kiAi=(1+Ak/2)∑k/2iAi,不断二分就可以 代码: #include <cstdio> #include <cstring> const int N = 45; int n, k; struct mat { int v[N][N]; mat() {memset(v, 0, sizeof(v));} mat operator * (ma…