读入一个自然数n,计算其各位数字之和,用汉语拼音写出和的每一位数字。
输入格式:每个测试输入包含1个测试用例,即给出自然数n的值。这里保证n小于10100。
输出格式:在一行内输出n的各位数字之和的每一位,拼音数字间有1 空格,但一行中最后一个拼音数字后没有空格。
输入样例:1234567890987654321123456789输出样例:
yi san wu
n小于10^100,那么他们的和肯定是三位数以内啦,这样就很简单了。
我们先按字符串输入,然后将每个数字保存到数组里面,求和就可以啦,然后计算个十百位分别是什么数字,输出就好了...
很常规的解法,如果大家有更好的方法欢迎讨论^_^
AC代码:
#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; }