读入一个正整数 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 10^100 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
第一种方法：栈
#include<string> #include<stack> #include<iostream> using namespace std; int main() { int sum = 0; string s; cin >> s; for (unsigned int i = 0; i < s.length(); i++) { sum += s[i] - '0'; } stack<string> PinYin; while (sum) { int digit = sum % 10; switch(digit) { case 0: PinYin.push("ling");break; case 1: PinYin.push("yi");break; case 2: PinYin.push("er");break; case 3: PinYin.push("san");break; case 4: PinYin.push("si");break; case 5: PinYin.push("wu");break; case 6: PinYin.push("liu");break; case 7: PinYin.push("qi");break; case 8: PinYin.push("ba");break; case 9: PinYin.push("jiu");break; default: break; } sum /= 10; } for (unsigned int i = 0; i <= PinYin.size() + 1; i++) { if (i == PinYin.size() + 1) { cout << PinYin.top(); } else { cout << PinYin.top() << " "; } PinYin.pop(); } return 0; }第二种方法：二维字符数组
#include<cstdio> #include<cstring> int main() { char qtc[110]; gets(qtc); int len = strlen(qtc); int sum = 0; for (int i = 0; i < len; i++) { sum += qtc[i] - '0'; // 将字符转化为数字并累加 } int num = 0, ans[10]; while (sum) { ans[num++] = sum % 10; // 将 sum中的每一位保存在数组中 sum /= 10; } char PinYin[10][5] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"}; for (int i = num - 1; i >= 0; i--) { printf("%s", PinYin[ans[i]]); // ans[i]即为高位开始的第i个数字 if (i) printf(" "); // i不为0则空格隔开 else printf("\n"); } return 0; }第三种方法： switch case
#include<stdio.h> #include<string.h> int main() { int sum = 0; char ch; ch = getchar(); while (ch != '\n') { switch (ch) { case '1': sum += 1; break; case '2': sum += 2; break; case '3': sum += 3; break; case '4': sum += 4; break; case '5': sum += 5; break; case '6': sum += 6; break; case '7': sum += 7; break; case '8': sum += 8; break; case '9': sum += 9; break; default: sum += 0; break; } ch = getchar(); //是继续往下读字符啦 } char *result[100]; int digit_count = 0; while(sum != 0) { int digit = sum % 10; switch (digit) { case 0: result[digit_count] = "ling"; break; case 1: result[digit_count] = "yi"; break; case 2: result[digit_count] = "er"; break; case 3: result[digit_count] = "san"; break; case 4: result[digit_count] = "si"; break; case 5: result[digit_count] = "wu"; break; case 6: result[digit_count] = "liu"; break; case 7: result[digit_count] = "qi"; break; case 8: result[digit_count] = "ba"; break; case 9: result[digit_count] = "jiu"; break; default: break; } sum = sum / 10; digit_count++; } for (int i = digit_count - 1; i >= 0; i--) { if (i == 0) { printf("%s", result[i]); } else { printf("%s ", result[i]); } } return 0; }