Project Euler Problem 36
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我是用了纯暴力枚举,枚举每个十进制回文数,并进行判断是否满足二进制回文,若满足则记录,方法很简单,代码如下:
#include<stdio.h>int bin[20];int sum;bool ispalindromic(int pali)//判断一个数是否在2进制数上循环{int j,n=pali;if(pali==0)return false;int top=0;while(pali){bin[top++]=pali%2;pali/=2;}for(j=0;j<top-j-1;j++)if(bin[j]!=bin[top-j-1])break;if(j>=top-j-1){sum+=n;return true;}return false;}int main(){int i,j,top,k,pali;sum+=0+1+3+5+7+9;//先计算个位数的情况for(i=1;i<=999;i+=2)//枚举abccba,aa,abba的情况,考虑到bc均能取到0{if(i<10){pali=i*10+i;if(ispalindromic(pali))printf("%d\n",pali);}if(i<100){pali=i+i/10*100+i%10*1000; if(ispalindromic(pali))printf("%d\n",pali);}pali=i+i/100*1000+(i%100)/10*10000+i%10*100000;if(ispalindromic(pali))printf("%d\n",pali);}for(i=1;i<=99;i+=2)//枚举aba,abcba的情况,考虑到bc均能取到0{for(k=0;k<=9;k++){if(i<10){pali=i*100+k*10+i;if(ispalindromic(pali))printf("%d\n",pali);}pali=i+i/10*1000+i%10*10000+k*100;if(ispalindromic(pali))printf("%d\n",pali);}}printf("%d\n",sum);return 0;}
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