python/TangDou/LanQiaoBei/ZhongGaoJi/复习1.md

复习提纲

以$A$为例，5进制理解为5^0,5^1,5^2,5^3 (1234)_5=1*4+3*5+2*25+1*125=194 (302)_8=2*8^0+0*8^1+3*8^2=2*1+0*8+3*64=194 (11000100)_2=(304)_8 与$(302)_8$不同，所以答案是C (C2)_{16}=2*1+12*16=194

#include <bits/stdc++.h>
using namespace std;

bool check(int x) {
    int sum = 0;
    while (x) {
        int a = x % 10;
        x /= 10;
        sum += a;
    }
    return sum % 2 == 0;
}

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = n; i <= m; i++)
        if (check(i))
            cout << i << endl;
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

int main() {
    string s;
    cin >> s;

    int m;
    cin >> m;
    m--;

    while (m--) {
        string t;
        for (int i = 0; i < s.size(); i++) {
            int c = 1;
            while (i + 1 < s.size() && s[i + 1] == s[i]) {
                c++;
                i++;
            }
            t = t + to_string(c) + s[i];
        }
        s = t;
    }
    cout << s << endl;
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int k;
string kill(string s) {
    int p = s[s.size() - 1]; // 最后一位是默认值
    for (int i = 0; i < s.size() - 1; i++)
        if (s[i + 1] > s[i]) {
            p = i;
            break;
        }
    string res;
    for (int i = 0; i < s.size(); i++)
        if (i != p) res += s[i];
    return res;
}

int main() {
    string s;
    cin >> s >> k;
    for (int i = 1; i <= k; i++) s = kill(s);
    cout << s << endl;
    return 0;
}