USACO Section 1.2.5 Palindromic Squares
题目
| <center style="font-size:14px;"> Palindromic Squares Rob Kolstad </center> 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: palsquareINPUT FORMATA single line with B, the base (specified in base 10).SAMPLE INPUT (file palsquare.in)10 OUTPUT FORMATLines 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
其实这个题目很简单，可是为毛我做了好久呢？其实，主要是我一贯试用cpp的string，这会想练练纯c，突然发现这玩意好繁琐。。又加上边聊边写，于是乎一道水题搞了一个多小时。
学习：
malloc(sizeof(char)*100); // 不清零
calloc(100,sizeof(char)); // 清零
然后，没了。
代码
#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;
}
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