Palindromic Squares
来源:互联网 发布:剑三花哥捏脸数据2017 编辑:程序博客网 时间:2024/05/21 20:24
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 12 43 911 12122 48426 676101 10201111 12321121 14641202 40804212 44944264 69696
#include <fstream>using namespace std;int main(int argc, const char * argv[]){ ofstream fout("palsquare.out"); ifstream fin("palsquare.in"); int zs_hexConversion(const int& num,const int &zs_hex,char *zsNum); bool zs_isPalindrome(int num,const char* zsNum); char zs_NUM[2048]; char zs[1024]; int zs_hex; fin>>zs_hex; for(int i=1;i<=300;i++) { if(zs_isPalindrome(zs_hexConversion(i*i,zs_hex,zs_NUM),zs_NUM)) { int length = zs_hexConversion(i,zs_hex,zs); length--; for(;length>=0;length--) fout<<zs[length]; fout<<" "<<zs_NUM<<endl; } } return 0;}int zs_hexConversion(const int& num,const int &zs_hex,char *zsNum){ int tempN; int _num = num; int i=0; for (;_num!=0;i++) { tempN = _num%zs_hex; _num=_num/zs_hex; if(tempN>=10) { tempN = tempN - 10; zsNum[i]=('A'+tempN); } else { zsNum[i]=('0'+tempN); } } zsNum[i]='\0'; return i;}bool zs_isPalindrome(int num,const char* zsNum){ num--; for(int i=0;i<num;i++,num--) { if(zsNum[i]!=zsNum[num]) return false; } return true;}
这道题唯一的知识点就是数制的转换。
思路:好像没什么难的,主要就是考进制转换,以及回文数的判断。这里要注意,最大的20进制中19表示为J,不要只CASE到15哦!
穷举1~300的所有平方数,转进制,比较,OK了,除非你不会怎么转进制。短除,然后逆序输出。
结果:
USER: jim zhai [jszhais1]TASK: palsquareLANG: C++Compiling...Compile: OKExecuting... Test 1: TEST OK [0.000 secs, 3356 KB] Test 2: TEST OK [0.000 secs, 3356 KB] Test 3: TEST OK [0.000 secs, 3356 KB] Test 4: TEST OK [0.000 secs, 3356 KB] Test 5: TEST OK [0.000 secs, 3356 KB] Test 6: TEST OK [0.000 secs, 3356 KB] Test 7: TEST OK [0.000 secs, 3356 KB] Test 8: TEST OK [0.000 secs, 3356 KB]All tests OK.Your program ('palsquare') produced all correct answers! This is yoursubmission #2 for this problem. Congratulations!
Here are the test data inputs:
------- test 1 ----10------- test 2 ----2------- test 3 ----5------- test 4 ----11------- test 5 ----15------- test 6 ----18------- test 7 ----20------- test 8 ----3Keep up the good work!
#include <stdio.h>#include <stdlib.h>#include <string.h>#include <assert.h>#include <ctype.h>#include <math.h>/* is string s a palindrome? */intispal(char *s){ char *t; t = s+strlen(s)-1; for(t=s+strlen(s)-1; s<t; s++, t--) if(*s != *t) return 0; return 1;}/* put the base b representation of n into s: 0 is represented by "" */voidnumbconv(char *s, int n, int b){ int len; if(n == 0) {strcpy(s, "");return; } /* figure out first n-1 digits */ numbconv(s, n/b, b); /* add last digit */ len = strlen(s); s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n%b]; s[len+1] = '\0';}voidmain(void){ char s[20]; char t[20]; int i, base; FILE *fin, *fout; fin = fopen("palsquare.in", "r"); fout = fopen("palsquare.out", "w"); assert(fin != NULL && fout != NULL); fscanf(fin, "%d", &base); for(i=1; i <= 300; i++) {numbconv(s, i*i, base);if(ispal(s)) { numbconv(t, i, base); fprintf(fout, "%s %s\n", t, s);} } exit(0);}
- Palindromic Squares
- palindromic squares
- Palindromic Squares
- Palindromic Squares
- Palindromic Squares
- Palindromic Squares
- Palindromic Squares
- Palindromic Squares
- Palindromic Squares
- Palindromic Squares(USACO)
- USACO 1.2-Palindromic Squares
- 1.2.4 Palindromic Squares
- Section 1.2 Palindromic Squares
- usaco 1.2:Palindromic Squares
- [USACO] Palindromic Squares
- usaco Palindromic Squares
- USACO 1.2 Palindromic Squares
- USACO Palindromic Squares
- 腾讯2012实习生笔试题2+答案解析
- ProxoolListener
- debian6.0安装之旅
- java RSA 不对称加密密钥生成、加密解密实例
- NYOJ 一个简单的数学题 南工330
- Palindromic Squares
- 腾讯2011年10月15日校招笔试+答案解析
- HDU 1079 Calendar Game
- DirectX9初步
- debian6安装以后无线网卡不能上网
- 操作Excel的经典类
- 腾讯2012实习生笔试题+答案解析
- 编程之美第一章游戏之乐学习总结
- ACM Steps_Chapter One_Section1