USACO S1.5 PrimePalindromes

題目大意:給定a,b(5<=a,b<=100,000,000),輸出[a,b]區間的迴文素數.

思路:先找出區間的迴文數.因爲以偶數或5開頭的,或位數爲偶數的迴文數(除11外)的不是素數.所以查找回文數時剪枝.

再對查找的迴文數判斷是否爲素數.可先用篩法篩出sqrt(b)以內素數.

 

可參考http://www.nocow.cn/index.php/USACO/pprime

 

/* ID:sheshao1 LANG:C++ TASK:pprime */ #include<cstdio> #include<cstring> #include<cmath> bool isprime[10002]; int prime[1300]; int total; int palind[1000]={2,3,5,7,11}; int len=5; int a,b; int get_prime(int n) { memset(isprime,1,sizeof(isprime)); int i,j; total=0; for(i=2;i<=n;i++){ if(isprime[i]) prime[total++]=i; for(j=0;j<total && i*prime[j]<=n;j++) { isprime[i*prime[j]]=0; if(i%prime[j]==0) break; } } return total; } bool judge(int n) { int i; for(i=0;i<total && prime[i]*prime[i]<=n;i++) if(n%prime[i]==0) return false; return true; } void get_palind() { int p[]={1,3,7,9}; int d[]={1,10,100,1000}; int i,j,k,l,t,tmp; for(i=1;i<4;i++) for(j=0;j<4;j++){ for(k=0,t=p[j]*d[i];k<d[i];k++,t++){ tmp=t*d[i]; for(l=i;l>0;l--) tmp+=((t/d[l])%10)*d[i-l]; if(tmp>b) return ; if(judge(tmp)) palind[len++]=tmp; } } } int main() { FILE *fin,*fout; int i; fin = fopen("pprime.in","r"); fout = fopen("pprime.out","w"); fscanf(fin,"%d%d",&a,&b); get_prime(int(sqrt(double(b)))+1); get_palind(); for(i=0;i<len;i++) if(palind[i]>=a && palind[i]<=b) fprintf(fout,"%d/n",palind[i]); return 0; } /* Executing... Test 1: TEST OK [0.000 secs, 3036 KB] Test 2: TEST OK [0.000 secs, 3036 KB] Test 3: TEST OK [0.022 secs, 3036 KB] Test 4: TEST OK [0.000 secs, 3036 KB] Test 5: TEST OK [0.011 secs, 3036 KB] Test 6: TEST OK [0.022 secs, 3036 KB] Test 7: TEST OK [0.022 secs, 3036 KB] Test 8: TEST OK [0.011 secs, 3036 KB] Test 9: TEST OK [0.011 secs, 3036 KB] */