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; }转载于:https://www.cnblogs.com/Dup4/p/9433198.html
