PAT 甲级 1005. Spell It Right (20) 【字符串】

最新推荐文章于 2024-11-12 15:59:45 发布

Dup4

最新推荐文章于 2024-11-12 15:59:45 发布

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分类专栏： PAT甲级文章标签：字符串

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本文提供了一种解决 PAT A 1005 的有效方法，使用字符串处理来应对大数问题，并通过 map 存储数字到英文单词的映射，最后输出计算结果的每位英文表示。

题目链接

https://www.patest.cn/contests/pat-a-practise/1005

思路

因为 n <= 10^100

所以要用字符串读入

但是 100 * 100 = 10000

所以 sum 用int 保存就好了
再把 sum 的每一位用英文输出

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;

string tran(int x)
{
    string s = "";
    if (x == 0)
        return "0";
    while (x)
    {
        s += x % 10 + '0';
        x /= 10;
    }
    return s;
}

int main()
{
    string s;
    cin >> s;
    int len = s.size();
    int sum = 0;
    for (int i = 0; i < len; i++)
        sum += s[i] - '0';
    map <char, string> m;
    m['0'] = "zero";
    m['1'] = "one";
    m['2'] = "two";
    m['3'] = "three";
    m['4'] = "four";
    m['5'] = "five";
    m['6'] = "six";
    m['7'] = "seven";
    m['8'] = "eight";
    m['9'] = "nine";
    string ans = tran(sum);
    len = ans.size();
    for (int i = len - 1; i >= 0; i--)
    {
        if (i != len - 1)
            printf(" ");
        cout << m[ans[i]];
    }
    cout << endl;
}