求更简单做法
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求更简单做法
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Sergei_Rachmaninoff楼主2022/12/14 09:12
#include <iostream>
using namespace std;
string n;
int main()
{
    cin >> n;
    int len = n.length();
    switch (len)
    {
    case 1:
        cout << n[0] - '0';
        break;
    case 2:
        cout << 9;
        break;
    case 3:
        cout << 9 + (n[0] - '0') * 100 + (n[1] - '0') * 10 + (n[2] - '0') - 100 + 1;
        break;
    case 4:
        cout << 900 + 9;
        break;
    case 5:
        cout << 900 + 9 + (n[0] - '0') * 10000 + (n[1] - '0') * 1000 + (n[2] - '0') * 100 + (n[3] - '0') * 10 + (n[4] - '0')-10000 + 1;
        break;
    case 6:
        cout << 90000 + 900 + 9;
        break;
    }
    return 0;
}
2022/12/14 09:12
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