POJ 2635 The Embarrassed Cryptographer [高精度求余+同余模定理]【数论】

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POJ2635-The Embarrassed Cryptographer

转载请注明出处：優YoU http://user.qzone.qq.com/289065406/blog/1309305108

大致题意：

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。

解题思路：

首先对题目的插图表示无语。。。

高精度求模+同余模定理

1、 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][890]。

为了方便处理，我的程序是按“局部有序，全局倒序”模式存放Kt

即Kt=[890][567][234][1 ] (一个中括号代表一个数组元素)

2、 素数打表，把10^6内的素数全部预打表，在求模时则枚举到小于L为止。

注意打表不能只打到100W，要保证素数表中最大的素数必须大于10^6，否则当L=100W且K为GOOD时，会因为数组越界而RE，这是因为越界后prime都是负无穷的数，枚举的while(prime[pMin]

//Memory Time
//624K  1235MS

#include<iostream>
#include<string.h>
using namespace std;

const int Range=1000100;  //打表不能只打到100W，素数表中最大的素数必须大于10^6

int Kt[10000];  //千进制的K
int L;
int prime[Range+1];

/*素数组打表*/
void PrimeTable(void)
{
int pNum=0;
prime[pNum++]=2;

for(int i=3;i<=Range;i+=2)  //奇偶法
{
bool flag=true;
for(int j=0;prime[j]*prime[j]<=i;j++)  //根号法+递归法
if(!(i%prime[j]))
{
flag=false;
break;
}
if(flag)
prime[pNum++]=i;
}
return;
}

/*高精度K对p求模，因数检查(整除)*/
bool mod(const int* K,const int p,const int len)
{
int sq=0;
for(int i=len-1;i>=0;i--)  //千进制K是逆序存放
sq=(sq*1000+K[i])%p;  //同余模定理

if(!sq)   //K被整除
return false;
return true;
}

int main(void)
{
PrimeTable();

char K[10000];
while(cin>>K>>L && L)
{
memset(Kt,0,sizeof(Kt));
int lenK=strlen(K);
for(int i=0;i<lenK;i++)  //把K转换为千进制Kt，其中Kt局部顺序，全局倒序
{                      //如K=1234567=[  1][234][567] ，则Kt=[567][234][1  ]
int pKt=(lenK+2-i)/3-1;
Kt[pKt]=Kt[pKt]*10+(K[i]-'0');
}
int lenKt=(lenK+2)/3;

bool flag=true;
int pMin=0;  //能整除K且比L小的在prime中的最小素数下标
while(prime[pMin]<L)  //枚举prime中比L小的素数
{
if(!mod(Kt,prime[pMin],lenKt))
{
flag=false;
cout<<"BAD "<<prime[pMin]<<endl;
break;
}
pMin++;
}
if(flag)
cout<<"GOOD"<<endl;
}
return 0;
}

按 Ctrl+C 复制代码

Sample Input

143 10

143 20

667 20

667 30

2573 30

2573 40

4 2

6 3

6 3

15 3

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 2

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 3

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 999981

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 999982

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 999983

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 999984

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999536689 999985

9936798836621706335903766366605021199756127575438907144689843371764114998372849970522970722679648297 1000000

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999924165887 1000000

9999999999999999997709341477512928270733515750111494296807693217401592660013176273247584305454312971 1000000

9999999999988881245087379264540384030358544520360773252628174690915590034078934845096473005364364269 1000000

9999999999999999999999999999999999999999999999999999999999999999999997947710886296926452585995644787 1000000

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998743929569 1000000

9999999999999999999999999999999999999999999999999999999999999999999999996406876316697599258447653751 1000000

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995271511 1000000

9999664515006205757944572422495695942633452678405393581216966782816097132509526872495414067984894021 1000000

0 0

Sample Output

GOOD

BAD 11

GOOD

BAD 23

GOOD

BAD 31

GOOD

BAD 2

BAD 2

GOOD

GOOD

GOOD

GOOD

GOOD

GOOD

BAD 999983

BAD 999983

BAD 587

BAD 100043

GOOD

GOOD

GOOD

GOOD

GOOD

BAD 16603

BAD 9103