USACO Section 1.2.5 Palindromic Squares

題目

Palindromic Squares Rob Kolstad Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome. Given a number base B (2 <= B <= 20 base 10), print all the integers N (1 <= N <= 300 base 10) such that the square of N is palindromic when expressed in base B; also print the value of that palindromic square. Use the letters 'A', 'B', and so on to represent the digits 10, 11, and so on. Print both the number and its square in base B. PROGRAM NAME: palsquare INPUT FORMAT A single line with B, the base (specified in base 10). SAMPLE INPUT (file palsquare.in) 10 OUTPUT FORMAT Lines with two integers represented in base B. The first integer is the number whose square is palindromic; the second integer is the square itself. SAMPLE OUTPUT (file palsquare.out) 1 1 2 4 3 9 11 121 22 484 26 676 101 10201 111 12321 121 14641 202 40804 212 44944 264 69696

思路

其實這個題目很簡單，可是為毛我做了好久呢？其實，主要是我一貫試用cpp的string，這會想練練純c，突然發現這玩意好繁瑣。。又加上邊聊邊寫，于是乎一道水題搞了一個多小時。

學習： malloc(sizeof(char)*100); // 不清零 calloc(100,sizeof(char)); // 清零 然後，沒了。

代碼

#include <cstdio>
#include <cstring>
#include <cstdlib>
const int dr[22]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K'};
const int N=300;
int n;
char * mir(char *a){
	int l=strlen(a);
	char *b=(char*)calloc(22,sizeof(char));
	for (int i(0);i<l;i++) b[i]=a[l-i-1];
	b[l]='\0';
	return b;
}
char * base(int k,int n){
	char *s=(char*)calloc(22,sizeof(char));
	while (k){
		int l=strlen(s);
		s[l]=dr[k%n];
		s[l+1]='\0';
		k/=n;
	}
	return mir(s);
}
int main(){
	freopen("palsquare.in","r",stdin);
	freopen("palsquare.out","w",stdout);
	scanf("%d",&n);
	for (int i(1);i<=N;i++){
		char *b1=base(i,n);
		char *b2=base(i*i,n);
		if (!strcmp(b2,mir(b2))){
		   printf("%s %s\n",b1,b2);
		}
	}
	return 0;
}