Palindromic Squares (USACO)
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原题地址
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 12 43 911 12122 48426 676101 10201111 12321121 14641202 40804212 44944264 69696
//这个是我的代码,USACO的题目都比较有技巧性/*ID : liuming9LANG : C++TASK: palsquare*/#include <cstdio>#include <algorithm>using namespace std;int n, size;char s[100], str[10];void palsquare(int a){int r;int x = a * a;size = 0;//把平方转化为n进制while(x){r = x % n;x /= n;if(0 <= r && r <= 9)s[size++] = r + '0';elses[size++] = r + 'A' - 10;}s[size] = 0;for(int i = 0; i < size / 2; i++)if(s[i] != s[size - 1 - i])return;size = 0;//把原数字转化为n进制while(a){r = a % n;a /= n;if(0 <= r && r <= 9)str[size++] = r + '0';elsestr[size++] = r + 'A' - 10;}str[size] = 0;//把原数字转化的n进制逆置for(int i = 0; i < size / 2; i++)swap(str[i], str[size - i - 1]);printf("%s %s\n", str, s);}int main(){freopen("palsquare.in", "r", stdin);freopen("palsquare.out", "w", stdout);scanf("%d", &n);for(int i = 1; i <= 300; i++){palsquare(i);}return 0;}
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