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;
}