大数同余（poj-2635）

（t*10）% k 会tle
转换为千进制

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e6 + 10000;
typedef long long ll;
int prime[maxn], isprime[maxn];
int cnt;
void get_prime(){
	cnt = 0;
	for(int i = 2; i < maxn; ++ i){
		if(isprime[i] == 0)
			prime[cnt ++] = i;
		for(int j = 0; j < cnt && prime[j] * i < maxn; ++ j){
			isprime[prime[j] * i] = 1;
			if(i % prime[j] == 0)
				break;
		}
	}
}

char s[110];
int a[50];
int l;
int main(){
	get_prime();
	while(~scanf("%s %d", s, &l)){
		if(s[0] == '0' && l == 0)
			break;
		int flag = 1, len = strlen(s), ans = 0, num = 0, temp = 0;
		for(int i = 0; i < len; ++ i){
			temp ++;
			if(temp == 3){ 
				a[num++] = s[i]-48 + (s[i-1]-48)*10 + (s[i-2]-48)*100;
				temp = 0;
			}
		}
		if(temp == 1)
			a[num++] = s[len-1]-48;
		else if(temp == 2)
			a[num++] = s[len-1]-48 + (s[len-2]-48)*10;
		for(int i = 0; i < cnt; ++ i){
			if(prime[i] >= l)
				break;
			int t = 0;
			for(int j = 0; j < num-1; ++ j){
				t = (t * 1000 + a[j]) % prime[i];
			}
			if(temp == 1)
				t = (t * 10 + a[num-1]) % prime[i];
			else if(temp == 2)
				t = (t * 100 + a[num-1]) % prime[i];
			else
				t = (t * 1000 + a[num-1]) % prime[i];
			if(t == 0){
				ans = prime[i];
				flag = 0;
				break;
			}
		}
		if(flag)
			puts("GOOD");
		else
			printf("BAD %d\n", ans);
	}
}