1023 Have Fun with Numbers
part 3, 3.1
自己解法
- 理解题意
- permutation：排列
- 注意溢出照样输出
#include <iostream>
using namespace std;
#include <algorithm>int main()
{string n, m = "", x;cin >> n;string c = n;int s = 0, d, flag = 1;for (int i = n.size() - 1; i >= 0; i--){d = (n[i] - '0') * 2 + s;s = d / 10;x = to_string(d % 10);if (flag == 0 || c.find(x) == string::npos)flag = 0;elsec[c.find(x)] = ' ';m.append(x);}if (s != 0)m.append(to_string(s));reverse(m.begin(), m.end());if (flag == 1)cout << "Yes" << endl;elsecout << "No" << endl;cout << m << endl;system("pause");return 0;
}
大神解法
- 柳神
#include <cstdio>
#include <string.h>
using namespace std;
int book[10];
int main() {char num[22];scanf("%s", num);int flag = 0, len = strlen(num);for(int i = len - 1; i >= 0; i--) {int temp = num[i] - '0';book[temp]++;temp = temp * 2 + flag;flag = 0;if(temp >= 10) {temp = temp - 10;flag = 1;}num[i] = (temp + '0');book[temp]--;}int flag1 = 0;for(int i = 0; i < 10; i++) {if(book[i] != 0)flag1 = 1;}printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n");if(flag == 1) printf("1");printf("%s", num);return 0;
}