Upside down primes
Time Limit: 2000ms
Memory Limit: 262144KB
This problem will be judged on CodeForcesGym. Original ID: 100753K64-bit integer IO format: %I64d Java class name: (Any)
解题：大数素性测试
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 LL mul(LL a,LL b,LL mod) { 5 if(!a) return 0; 6 return ((a&1)*b%mod + (mul(a>>1,b,mod)<<1)%mod)%mod; 7 } 8 LL quickPow(LL a,LL d,LL n){ 9 LL ret = 1; 10 while(d){ 11 if(d&1) ret = mul(ret,a,n); 12 d >>= 1; 13 a = mul(a,a,n); 14 } 15 return ret; 16 } 17 bool check(LL a,LL d,LL n){ 18 if(n == a) return true; 19 while(~d&1) d >>= 1; 20 LL t = quickPow(a,d,n); 21 while(d < n-1 && t != 1 && t != n-1){ 22 t = mul(t,t,n); 23 d <<= 1; 24 } 25 return (d&1)||t == n-1; 26 } 27 bool isP(LL n){ 28 if(n == 2) return true; 29 if(n < 2 || 0 == (n&1)) return false; 30 static int p[10] = {2,3,5,7,11,61,24251}; 31 for(int i = 0; i < 7; ++i) 32 if(!check(p[i],n-1,n)) return false; 33 return true; 34 } 35 36 bool trans(LL x){ 37 char str[100]; 38 sprintf(str,"%I64d",x); 39 for(int i = 0; str[i]; ++i) 40 if(str[i] == '3' || str[i] == '4' || str[i] == '7') return false; 41 reverse(str,str + strlen(str)); 42 for(int i = 0; str[i]; ++i) 43 if(str[i] == '6') str[i] = '9'; 44 else if(str[i] == '9') str[i] = '6'; 45 sscanf(str,"%I64d",&x); 46 return isP(x); 47 } 48 int main(){ 49 LL x; 50 while(~scanf("%I64d",&x)){ 51 if(isP(x) && trans(x)) puts("yes"); 52 else puts("no"); 53 } 54 return 0; 55 }