poj2635The Embarrassed Cryptographer(数论)(解题报告)
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http://poj.org/problem?id=2635
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The Embarrassed Cryptographer
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 10080
Accepted: 2639
Description
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 to be secure because there is no known method for factoring such a product effectively.
What Odd Even did not think of, was that both factors in a key should be large, not just their product. It is now possible that some of the users of the system have weak keys. In a desperate attempt not to be fired, Odd Even secretly goes through all the users keys, to check if they are strong enough. He uses his very poweful Atari, and is especially careful when checking his boss' key.
Input
The input consists of no more than 20 test cases. Each test case is a line with the integers 4 <= K <= 10100 and 2 <= L <= 106. K is the key itself, a product of two primes. L is the wanted minimum size of the factors in the key. The input set is terminated by a case where K = 0 and L = 0.
Output
For each number K, if one of its factors are strictly less than the required L, your program should output "BAD p", where p is the smallest factor in K. Otherwise, it should output "GOOD". Cases should be separated by a line-break.
Sample Input
143 10
143 20
667 20
667 30
2573 30
2573 40
0 0
Sample Output
GOOD
BAD 11
GOOD
BAD 23
GOOD
BAD 31
Source
Nordic 2005
题意:给出k和L,,要求K是由两个素因子组成,且K是由两个素因子乘积形成,要求其中一个素因子小于L
思路:高精度取模+同余定理。
1.构造素数表格,这里可用奇偶法,和线性素数筛选法
2,由于给出K非常大,因此可将K化为千进制存到一组数组中,以提高高精度运算的效率
//Memory 480K Time1266MS length1207B#include<stdio.h>#include<stdlib.h>#include<string.h>#include<iostream>using namespace std;const int maxn=1001000;int len,t2,total;int prime[maxn+1];int k[40];char ch[105];void makeprime()//构造素数表 { total=0; int flag; prime[total++]=2; for(int i=3;i<=maxn;i+=2) {flag=1; for(int j=0;prime[j]*prime[j]<=i;j++) {if(i%prime[j]==0) { flag=0; break; } } if(flag) prime[total++]=i; } }
void change()//把字符串转化为千进制 {int t1=len%3;t2=len/3;if(t1==0)t2--;int j=t2;memset(k,0,sizeof(k));for(int i=len-1;i>=0;i-=3)if(j>=0){for(int t=(i-2>0? i-2:0);t<=i;t++)k[j]=k[j]*10+ch[t]-'0';j--;}}int Mod(int d)//大数取模运算 {int ans=0;for(int i=0;i<=t2;i++)ans=(ans*1000+k[i])%d;return ans;}int main(){makeprime();int L;while(~scanf("%s%d",ch,&L)&&L) { len=strlen(ch); int ok=1; change(); for(int i=0;prime[i]<L&&i<total;i++) if(Mod(prime[i])==0) { printf("BAD %d\n",prime[i]); ok=0; break; } if(ok) printf("GOOD\n"); } system("pause"); return 0;}
小结:对于素数的筛选和构造中,可用三种方法:
1,用时最短的是线性筛选法,但耗空间;
2.奇偶法筛选选素数:省空间,但与线性筛选法相比,比较耗时。
3.暴力枚举,超时!!!
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