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http://poj.org/problem?id=2635
题意:给出一个大数K和一个数L,判断是否存在一个小于L的素数能够整除K,存在的话输出最小的那个素数
同余定理:1234%3可以分解为
1%3=1
(1*10+2)%3=0
(0*10+3)%3=0
(0*10+4)%3=1
则1234对3取余得1,大数亦如此,但是本题中需要将这个大数分为多个1000进制的数,10进制的话会TLE
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <stack>
#include <vector>
#include <map>
using namespace std;
#define N 1000200
#define INF 0xfffffff
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))
typedef long long LL;
int isprim[N] = {1, 1}, prim[N], a[N], k = 0;
void Init ()
{
for (int i=2; i<N; i++)
{
if (!isprim[i])
{
prim[k++] = i;
for (int j=i+i; j<N; j+=i)
isprim[j] = 1;
}
}
}
int main ()
{
char str[N];
int m;
Init();
while (scanf ("%s %d", str, &m), strcmp("0", str) || m)
{
met (a, 0);
int cnt = 0, len = strlen(str);
int i = len-1;
while (i>=0)
{///将字符串中的数以1000进制的形式倒着存入a数组中
///1 234 567存为a[567][234][1]
if (i-2>=0)
{
a[cnt] = (str[i-2]-'0')*100+(str[i-1]-'0')*10+str[i]-'0';
i -= 3;
}
else if (i-1>=0)
{
a[cnt] = (str[i-1]-'0')*10+str[i]-'0';
i -= 2;
}
else a[cnt] = str[i--]-'0';
cnt++;
}
int flag = 0;
for (i=0; i<k; i++)
{
if (prim[i] >= m) break;
LL mod = 0;
for (int j=cnt-1; j>=0; j--)
mod = (mod * 1000 + a[j]) % prim[i];
if (!mod)
{
printf ("BAD %d\n", prim[i]);
flag = 1;
break;
}
}
if (!flag) puts ("GOOD");
}
return 0;
}