题目链接:http://poj.org/problem?id=2635 题目分析: http://blog.csdn.net/lyy289065406/article/details/6648530…

The Embarrassed Cryptographer Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13041 Accepted: 3516 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of user…

题意:给出一个大数,这个大数由两个素数相乘得到,让我们判断是否其中一个素数比L要小,如果两个都小,输出较小的那个. 分析:大数求余的方法:针对题目中的样例,143 11,我们可以这样算,1 % 11 = 1:      1×10 + 4 % 11 = 3:      3×10 + 3 % 11 = 0;我们可以把大数拆成小数去计算,同余膜定理保证了这个算法的这正确性,而且我们将进制进行一定的扩大也是正确的. 注意:素数打标需要优化,否则超时.   进制需要适当,100和1000都可以,10进制超…

The Embarrassed Cryptographer Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11978   Accepted: 3194 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of…

The Embarrassed Cryptographer DescriptionThe young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in his company. The cryptographic keys are created from th…

The Embarrassed Cryptographer Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15069   Accepted: 4132 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of…

题目: http://poj.org/problem?id=2635 利用同余模定理大数拆分取模,但是耗时,需要转化为高进制,这样位数少,循环少,这里转化为1000进制的,如果转化为10000进制,需要long long #include <stdio.h> #include <stdlib.h> #include <string.h> #include <cmath> using namespace std; ]; ]; void prime_init()…

题意: 项的自幂级数求和为 11 + 22 + 33 + - + 1010 = 10405071317. 求如下一千项的自幂级数求和的最后10位数字:11 + 22 + 33 + - + 10001000. 思路: 求最后十位数字 % 1010 即可. 对于快速幂中数据溢出的问题,有两种解决方法: 1. 方法一:对于两个数 x y,现在想求 x * y % MOD,可以将 x 表示成 a * DIGS + b,y 表示成 c * DIGS + d,x * y % MOD 则等价与 ( a * c…

Large Division Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T…

Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T (≤ 525), denot…