UVA 401 Palindromes

本文介绍了一个C++程序,用于判断输入的字符串是否为回文或镜像字符串。程序使用了预处理宏定义简化循环及输入输出,并定义了一个映射表来检查镜像特性。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

输入一个字符串,判断它是否回文,还有镜像(mirrored),有可能两类都是或都不是。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<cmath>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 100000000;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 44;
const int M = 11;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

string s;
char m[255];

bool check1()
{
    int a = 0, b = s.size() - 1;
    while(a < b)
    {
        if(s[a] != s[b])return 0;
        a++;
        b--;
    }
    return 1;
}

bool check2()
{
    int a = 0, b = s.size() - 1;
    while(a <= b)
    {
        if(s[a] != m[s[b]])return 0;
        a++;
        b--;
    }
    return 1;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    CLR(m, 0);
    m['A'] = 'A';m['I'] = 'I';m['M'] = 'M';m['T'] = 'T';m['W'] = 'W';m['Z'] = '5';m['3'] = 'E';
    m['E'] = '3';m['J'] = 'L';m['O'] = 'O';m['U'] = 'U';m['X'] = 'X';m['1'] = '1';m['5'] = 'Z';
    m['H'] = 'H';m['L'] = 'J';m['S'] = '2';m['V'] = 'V';m['Y'] = 'Y';m['2'] = 'S';m['8'] = '8';
    int f1, f2;
    while(cin >> s)
    {
        f1 = check1();
        f2 = check2();
        cout << s << " -- is ";
        if(f1 && f2)puts("a mirrored palindrome.");
        else if(f1)puts("a regular palindrome.");
        else if(f2)puts("a mirrored string.");
        else puts("not a palindrome.");
        puts("");
    }
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值