讀入一個自然數n,計算其各位數字之和,用漢語拼音寫出和的每一位數字。
輸入格式:每個測試輸入包含1個測試用例,即給出自然數n的值。這裏保證n小於10100。
輸出格式:在一行內輸出n的各位數字之和的每一位,拼音數字間有1 空格,但一行中最後一個拼音數字後沒有空格。
輸入樣例:1234567890987654321123456789輸出樣例:
yi san wu
n小於10^100,那麼他們的和肯定是三位數以內啦,這樣就很簡單了。
我們先按字符串輸入,然後將每個數字保存到數組裏面,求和就可以啦,然後計算個十百位分別是什麼數字,輸出就好了...
很常規的解法,如果大家有更好的方法歡迎討論^_^
AC代碼:
#include <iostream>
#include <string>
using namespace std;
string chinese[10] = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };
int convert(string s)
{
int n,sum;
sum = 0;
n = s.size();
int *a = new int[n];
for (int i = 0; i < n; i++)
{
a[i] = s[i] - '0';
sum += a[i];
}
return sum;
}
int main()
{
string s;
int p[3];
cin >> s;
int sum;
int count = 0;
sum = convert(s);
int tmp;
tmp = sum;
while (tmp > 0)
{
tmp = tmp / 10;
count++;
}
switch (count)
{
case 1:
cout << chinese[count];
break;
case 2:
p[0] = sum / 10;
p[1] = sum % 10;
cout << chinese[p[0]] << " " << chinese[p[1]];
break;
default:
p[0] = sum / 100;
sum = sum % 100;
p[1] = sum / 10;
p[2] = sum % 10;
cout << chinese[p[0]] << " " << chinese[p[1]] << " "<<chinese[p[2]];
break;
}
return 0;
}