USACO 1.5 Prime Palindromes(枚举)
2012-11-02 16:53
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本来想筛选素数的。。。交了之后发现复杂度也太高了。。。可耻的看了一下hint,然后就各种for枚举了每一位了。.。
/* ID: cuizhe LANG: C++ TASK: pprime */ #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 1000001 bool p ; int prim[100000],num; int judge(int x) { int i; for(i = 1;i <= num-1;i ++) { if(prim[i] > x) break; if(x%prim[i] == 0) return 0; } return 1; } int main() { int i,j,k,u; long long t,a,b; freopen("pprime.in","r",stdin); freopen("pprime.out","w",stdout); scanf("%lld%lld",&a,&b); for(i = 2; i <= N; i ++) { if(!p[i]) { for(j = i+i; j <= N; j += i) { p[j] = 1; } } } num = 1; for(i = 2;i <= N;i ++) { if(!p[i]) prim[num++] = i; } for(i = 1;i <= 9;i += 2)//一位 { if(!p[i]&&i >= a&&i <= b) printf("%d\n",i); } for(i = 1;i <= 9;i += 2)//两位 { if(!p[i*10+i]&&i*10+i>=a&&i*10+i<=b) printf("%d\n",i*10+i); } for(i = 1;i <= 9;i += 2)//三位 { for(j = 0;j <= 9;j ++) { if(!p[i*100+j*10+i]&&i*100+j*10+i>=a&&i*100+j*10+i<=b) printf("%d\n",i*100+j*10+i); } } for(i = 1;i <= 9;i += 2)//四位 { for(j = 0;j <= 9;j ++) { if(!p[i*1000+j*100+10*j+i]&&i*1000+j*100+10*j+i>=a&&i*1000+j*100+10*j+i<=b) printf("%d\n",i*100+j*100+10*j+i); } } for(i = 1;i <= 9;i += 2)//五位 { for(j = 0;j <= 9;j ++) { for(k = 0;k <= 9;k ++) { t = i*10000+i+1000*j+10*j+k*100; if(!p[t]&&t>=a&&t<=b) printf("%lld\n",t); } } } for(i = 1;i <= 9;i += 2)//六位 { for(j = 0;j <= 9;j ++) { for(k = 0;k <= 9;k ++) { t = i*100000+i+10000*j+10*j+k*1000+k*100; if(!p[t]&&t >= a&&t <= b) printf("%lld\n",t); } } } for(i = 1;i <= 9;i += 2)//七位 { for(j = 0;j <= 9;j ++) { for(k = 0;k <= 9;k ++) { for(u = 0;u <= 9;u ++) { t = 1000000*i+i+100000*j+10*j+10000*k+k*100+1000*u; if(judge(t)&&t >= a&&t <= b) printf("%lld\n",t); } } } } for(i = 1;i <= 9;i += 2)//八位 { for(j = 0;j <= 9;j ++) { for(k = 0;k <= 9;k ++) { for(u = 0;u <= 9;u ++) { t = 10000000*i+i+1000000*j+10*j+100000*k+k*100+10000*u+1000*u; if(judge(t)&&t >= a&&t <= b) printf("%lld\n",t); } } } } return 0; }
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