The 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 the product of two primes, and are believed…

大数取MOD... The Embarrassed Cryptographer Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11359 Accepted: 3026 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousan…

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 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15767   Accepted: 4337 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of…

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…

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…

The Embarrassed Cryptographer 题意 给一个两个素数乘积(1e100)K, 给以个数L(1e6), 判断K的两个素数是不是都大于L 题解 对于这么大的范围,素数肯定是要打表(可采用埃筛,欧拉筛,莫比乌斯筛);这里有别人模板 简单的想法是遍历表中<L的素数去模K(即便对K分解也是如此办的) 但K很大需要高精度取模,由于足够大需要转换成K,L都足够大需要转换成千进制 代码 素数打表 int isp[maxn];// isp[i]=0 i是素数 int su[maxn];/…

题目:http://poj.org/problem?id=2635 高精度求模  同余模定理. 题意: 给定一个大数K,K是两个大素数的乘积的值.再给定一个int内的数L 问这两个大素数中最小的一个是否小于L,如果小于则输出这个素数. 思路: Char格式读入K.把K转成千进制Kt,同时变为int型. 把数字往大进制转换能够加快运算效率.若用十进制则耗费很多时间,会TLE.千进制的性质与十进制相似. 例如,把K=1234567890转成千进制,就变成了:Kt=[  1][234][567][89…

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()…