USACO——Palindromic Squares 回文平方数

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. NOTE WELL THAT BOTH INTEGERS ARE IN BASE B!

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：做个判断回文的函数和进制转化的函数即可

/*
ID: youqihe1
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include<string.h>

using namespace std;


int pal(char A[])
{
    int len=strlen(A),i,j;
    for(i=0;i<len/2;i++)
        if(A[i]!=A[len-i-1])
        return 0;
    return 1;
}
void quare(int n,int b,char C[])
{
    int i=0;
    char B[25];
    memset(B,'\0',sizeof(B));
    while(n)
    {
        int k=n%b;
        if(k<10)
            B[i++]=k+'0';
        else
        {
            if(k==10) B[i++]='A';
            if(k==11) B[i++]='B';
            if(k==12) B[i++]='C';
            if(k==13) B[i++]='D';
            if(k==14) B[i++]='E';
            if(k==15) B[i++]='F';
            if(k==16) B[i++]='G';
            if(k==17) B[i++]='H';
            if(k==18) B[i++]='I';
            if(k==19) B[i++]='J';
        }
        n/=b;
    }
    int t=0,j;
    for(j=i-1;j>=0;j--,t++)
        C[j]=B[t];
}

int main() {
    FILE *fin  = fopen ("palsquare.in", "r");
    FILE *fout = fopen ("palsquare.out", "w");
    int N;
    fscanf(fin,"%d",&N);
    int i,j;
    for(i=1;i<=300;i++)
    {
        char B[25],C[25];
        memset(B,'\0',sizeof(B));
        memset(C,'\0',sizeof(C));
        quare(i*i,N,B);
        if(pal(B))
        {
            quare(i,N,C);
//            printf("%d ",i);
//            printf("%s %s\n",C,B);
           fprintf(fout,"%s %s\n",C,B);
        }
    }
    return 0;
}

同样的，还有这题

Dual Palindromes
Mario Cruz (Colombia) & Hugo Rickeboer (Argentina)

A number that reads the same from right to left as when read from left to right is called a palindrome. The number 12321 is a palindrome; the number 77778 is not. Of course, palindromes have neither leading nor trailing zeroes, so 0220 is not a palindrome.

The number 21 (base 10) is not palindrome in base 10, but the number 21 (base 10) is, in fact, a palindrome in base 2 (10101).

Write a program that reads two numbers (expressed in base 10):

• N (1 <= N <= 15)
• S (0 < S < 10000)

and then finds and prints (in base 10) the first N numbers strictly greater than S that are palindromic when written in two or more number bases (2 <= base <= 10).

Solutions to this problem do not require manipulating integers larger than the standard 32 bits.

PROGRAM NAME: dualpal

INPUT FORMAT

A single line with space separated integers N and S.

SAMPLE INPUT (file dualpal.in)

3 25

OUTPUT FORMAT

N lines, each with a base 10 number that is palindromic when expressed in at least two of the bases 2..10. The numbers should be listed in order from smallest to largest.

SAMPLE OUTPUT (file dualpal.out)

26
27
28

也只要一个个判断即可

/*
ID: youqihe1
PROG: dualpal
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include<string.h>

using namespace std;
int pal(char A[])
{
    int len=strlen(A),i,j;
    for(i=0;i<len/2;i++)
        if(A[i]!=A[len-i-1])
        return 0;
    return 1;
}
void quare(int n,int b,char C[])
{
    int i=0;
    char B[30];
    memset(B,'\0',sizeof(B));
    while(n)
    {
        int k=n%b;
            B[i++]=k+'0';
        n/=b;
    }
    int t=0,j;
    for(j=i-1;j>=0;j--,t++)
        C[j]=B[t];
}
int main() {
    FILE *fin  = fopen ("dualpal.in", "r");
    FILE *fout = fopen ("dualpal.out", "w");
    int N,S;
    fscanf(fin,"%d %d",&N,&S);
    int i,j,k,num=0;
    for(i=S+1;num<N;i++)
    {
        int sum=0;
        for(k=2;k<=10;k++)
        {
            char B[30];
            memset(B,'\0',sizeof(B));
            quare(i,k,B);
            if(pal(B))
                sum++;
            if(sum==2)
            {
                fprintf(fout,"%d\n",i);
                num++;
                break;
            }
        }
    }


    return 0;
}