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10^6int Kt[10000];  //千进制的Kint 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;}

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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