USACODual Palindromes回文

<center style="font-family: Simsun;font-size:14px;"><strong><span style="font-size:48px;">Dual Palindromes</span>
Mario Cruz (Colombia) & Hugo Rickeboer (Argentina)</strong></center><p style="font-family: Simsun;font-size:14px;">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.</p><p style="font-family: Simsun;font-size:14px;">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).</p><p style="font-family: Simsun;font-size:14px;">Write a program that reads two numbers (expressed in base 10):</p><ul style="font-family: Simsun;font-size:14px;"><li>N (1 <= N <= 15)</li><li>S (0 < S < 10000)</li></ul><span style="font-family: Simsun;font-size:14px;">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).</span><p style="font-family: Simsun;font-size:14px;">Solutions to this problem do not require manipulating integers larger than the standard 32 bits.</p><h3 style="font-family: Simsun;">PROGRAM NAME: dualpal</h3><h3 style="font-family: Simsun;">INPUT FORMAT</h3><p style="font-family: Simsun;font-size:14px;">A single line with space separated integers N and S.</p><h3 style="font-family: Simsun;">SAMPLE INPUT (file dualpal.in)</h3><pre>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

<pre name="code" class="cpp">/*
TASK:dualpal
LANG:C++
ID:huibaochen
*/
/*给你n,s两个数，计算从s+1开始的数转换成2——10进制中，至少有两个或以上的数是回文数，符合条件的就输出，输出n个后结束程序。*/
#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;
char ch[100 + 5];

bool yes(int a, int b)
{
    int len = 0;
    memset(ch, 0, sizeof(ch));
    int r;
    while(a){           //进制转化
        r = a % b;
        a = a / b;
        ch[len++] = r >= 0 && r <= 9 ? r + '0' : r + 'A' - 10;
    }
    for(int i = 0; i < len; i++){           //判断是否为回文
        if(ch[i] != ch[len - 1 - i])
            return 0;
    }
    return 1;
}

int main()
{
    freopen ("dualpal.in", "r", stdin);
    freopen ("dualpal.out", "w", stdout);
    int n, s, num;
    scanf("%d%d", &n, &s);
    for(int i = s + 1; n > 0; i++){  //从s + 1开始
        num = 0;
        for(int j = 2; j <= 10 ; j++){   //转化为2——10进制
            if(yes(i, j))
                num++;
        }
        if(num >= 2){     //至少有两个符合条件，就输出
            printf("%d\n", i);
            n--;
        }
    }
    return 0;
}