训练3.19

1.H（简单模拟，注意细节）

#include<bits/stdc++.h>
#define int long long
using namespace std;
int er[30],q[1000000];
bool cmd(char a,char b,char c)
{
    if(a>='a'&&a<='z'&&c>='a'&&c<='z'&&a<c&&b=='-'||a>='0'&&a<='9'&&c>='0'&&c<='9'&&a<c&&b=='-')
        return true;
    else return false;
}
signed main()
{
    int p1,p2,p3;
    cin>>p1>>p2>>p3;
    string s;
    cin>>s;
    for(int i=0;i<s.size();i++)
    {
            if (cmd(s[i - 1], s[i],s[i + 1])) {
                if (p1 == 1 && p3 == 1) {
                    for (int j = s[i - 1] + 1; j < s[i + 1]; j++) {
                        for (int k = 0; k < p2; k++)
                            cout << (char) j;
                    }
                } else if (p1 == 2 && p3 == 1) {
                    for (int j = s[i - 1] + 1; j < s[i + 1]; j++) {
                        for (int k = 0; k < p2; k++) {
                            if (s[i - 1] >= 'a' && s[i - 1] <= 'z')cout << (char) (j - 32);
                            else cout << (char) j;
                        }
                    }
                } else if (p1 == 3) {
                    for (int j = s[i - 1] + 1; j < s[i + 1]; j++) {
                        for (int k = 0; k < p2; k++)
                            cout << '*';
                    }
                } else if (p1 == 1 && p3 == 2) {
                    for (int j = s[i + 1] - 1; j > s[i - 1]; j--) {
                        for (int k = 0; k < p2; k++)
                            cout << (char) j;
                    }
                } else if (p1 == 2 && p3 == 2) {
                    for (int j = s[i + 1] - 1; j > s[i - 1]; j--) {
                        for (int k = 0; k < p2; k++) {
                            if (s[i - 1] >= 'a' && s[i - 1] <= 'z')cout << (char) (j - 32);
                            else cout << (char) j;
                        }
                    }
                } else cout << "-";
            }
        else cout<<s[i];
    }
    return 0;
}

2.I（通过数字所在的位置来判断二进制串下一个是0或1，再倒过来，但要注意longlong会爆，最后判断初始位，输出即可）

#include<bits/stdc++.h>
#define int unsigned long long
using namespace std;
int a[70];
signed main()
{
    int n,k;
    cin>>n>>k;
    vector<int >v;
    int d;
    if (n == 64) {
        d = (1ull<<63)+(1ull<<63)-1ull;//用来取到2的64次方减一
    }
    else d = (1ull << n) - 1;
    while(d>1)
    {
        if(d/2>=k>0)
        {
            v.push_back(0);
            d/=2;
        }
        else
        {
            k=(d-k);
            v.push_back(1);
            d/=2;
        }
    }
    std::reverse(v.begin(), v.end());
    for(int i=0;i<v.size();i++)cout<<v[i];
    cout<<endl;
    string s;
    if (k == 0) {
        s = "0";
    } else s = "1";
    for (int i = 0; i < v.size(); ++i) {
        if (v[i] == 0) {
            s = '0' + s;
        } else s = '1' + s;
    }
    cout << s << endl;
    return 0;
}

3.J (结构体排序后，由于胜者和败者都是有序的数列，直接暴力不可行就利用归并排序，使用stl的merge函数排序即可)

#include<bits/stdc++.h>
using namespace std;
#define int  long long
struct init{
    int a,b,c;
};
bool cmp(init x,init y){
    if(x.b!=y.b)return x.b>y.b;
    return x.a<y.a;
}
signed main(){
    int n,r,q;
    cin>>n>>r>>q;
    n*=2;
    vector<init>a(n);
    for (int i = 0; i <n ; ++i)
    {
        cin>>a[i].b;
        a[i].a=i+1;
    }
    for (int i = 0; i <n ; ++i)
    {
        cin>>a[i].c;
    }
    sort(a.begin(), a.end(),cmp);
    while(r--){
        vector<init>p,q;
        for (int i = 0; i <n ; i+=2)
        {
            if(a[i].c>a[i+1].c){
                a[i].b++;
                p.push_back(a[i]);
                q.push_back(a[i+1]);
            }
            else
            {
                a[i+1].b++;
                p.push_back(a[i+1]);
                q.push_back(a[i]);
            }
        }
        merge(p.begin(),p.end(),q.begin(),q.end(),a.begin(),cmp);
    }
    cout<<a[q-1].a;
}