POJ -2635-The Embarrassed Cryptographer-数论

题意：

  给你一个数n，k，如果n有小于k的素数因子，就输出，没有的话输出GOOD

思路：

 

求一个10^100 的K 对 小于 10^6的 L 取余，问2-L-1 之间能否有能整除K的素数。

设k=a0a1a2a3a4a5a6a7a8a9a10.

k % m =a0a1a2a3a4a5a6*1000 %m + a7a8a9a10%m

          =(a0a1a2*1000 %m + a3a4a5a6)*1000 %m + a7a8a9a10%m

于是很明显可以递归求解，k%m

于是枚举m，每次求解k就好了。。。。

注意素数打表的时候用j=i+i！！！

还有如果用as==0判断结果的时候注意可能l太小（l==2）而不进入循环，导致as未初始化，

所以as要初始化为1.

CODE  1010ms

#include<stdio.h>#include<math.h>#include<stdlib.h>#include<string.h>#include<iostream>#define MAX 1000100using namespace std;int prime[MAX];bool ls[MAX];int top;void makeprime()///素数{    int i, j;    top = 0;    for(i = 2; i <MAX;i++)    {        if(!ls[i])        {            prime[top++] = i;            for(j =i+i;j < MAX;j+=i)                ls[j] = i;        }    }    prime[top] = MAX;}int main(){    makeprime();    char s[150];    int k;    while(~scanf("%s%d",s,&k)&&k)    {        int len = strlen(s);        int io = len%5;        int num = 0, i;        for(i = 0; i < io; i++)            num=(num+s[i]-'0')*10;            num/=10;            int as = 1,j;        for(j = 0;prime[j] < k; j++)///10000进制        {            int pm = prime[j];            as = num%pm;            for(i = io; i < len;i+=5)            {             as=((long long)as*100000+(s[i]-'0')*10000+(s[i+1]-'0')*1000+(s[i+2]-'0')*100+(s[i+3]-'0')*10 + (s[i+4]-'0'))%pm;            }            if(as == 0)            {                printf("BAD %d\n",pm);                break;            }        }        if(as)        {            printf("GOOD\n");        }    }    return 0;}

数据：

 

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