赛前集训 ——DAY 1

太痛苦了
cqr出题堪比李老师

今天改不动题，打了往年回放和一本通域的题库题

[CSP-J2020] 优秀的拆分
改后AC，一开始想的太复杂了，利用贪心思想就能AC了

#include<bits/stdc++.h>
using namespace std;
long long n,a[30];
int main() {
    freopen("power.in","r",stdin);
    freopen("power.out","w",stdout);
    cin>>n;
    if(n%2==1) {
        cout<<-1;
        return 0;
    }
    a[1]=1;
    for(int i=2;i<=26;i++)a[i]=a[i-1]*2;
    for(int i=26;i>=2;i--)
        if(n>=a[i]) {
            cout<<a[i]<<" ";
            n=n-a[i];
        }
    return 0;
}

[CSP-J 2021] 网络连接
仍在改进，感觉越打越乱，不知道错哪了

#include<bits/stdc++.h>
using namespace std;
int n,i,j,q=1,pos,ans,o,u;
bool f=true;
struct Computer {
    int num=0;
    string s1,s2;
    char s3[4];
} a[10001];

int main () {
    cin>>n;
    for(i=1; i<=n; i++) {
        cin>>a[i].s1>>a[i].s2;
        a[i].num=a[i-1].num+1;
    }

    for(i=1; i<=n; i++) {
        int c=1;
        for(j=1; j<=a[i].s2.size(); i++) {
            if(a[i].s2[j]==&#039;.&#039; && a[i].s2[j]==&#039;:&#039; && j==a[i].s2.size()) {
                if(a[i].s2[j]==&#039;.&#039;)ans++;
                if(a[i].s2[j]==&#039;:&#039;)pos++;
                if(j==a[i].s2.size()) {
                    if(a[i].s2[q]==&#039;0&#039;) {
                        a[i].s3[1]=&#039;E&#039;;
                        a[i].s3[2]=&#039;R&#039;;
                        a[i].s3[3]=&#039;R&#039;;
                        break;
                    }
                    /*统计字符个数，特判&#039;:&#039;之后*/

                    int v=j-q;
                    for(int op=v; op>=1; op--) {
                        o += (int)a[i].s2[q] * v;
                    }
                    if(o>65535 || o<0) {
                        a[i].s3[1]=&#039;E&#039;;
                        a[i].s3[2]=&#039;R&#039;;
                        a[i].s3[3]=&#039;R&#039;;
                        break;
                    }
                    //判断&#039;:&#039;之后

                    q=j+1;
                }
                if(a[i].s2[q]==&#039;0&#039;) {
                    a[i].s3[1]=&#039;E&#039;;
                    a[i].s3[2]=&#039;R&#039;;
                    a[i].s3[3]=&#039;R&#039;;
                    break;
                }
                //特判&#039;.&#039;与&#039;:&#039;之前

                int v=j-q;
                for(int op=v; op>=1; op--) {
                    u += (int)a[i].s2[q] * v;
                }
                if(u>255 || u<0) {
                    a[i].s3[1]=&#039;E&#039;;
                    a[i].s3[2]=&#039;R&#039;;
                    a[i].s3[3]=&#039;R&#039;;
                    break;
                }
                //判断&#039;.&#039;与&#039;:&#039;之前

                q=j+1;
            }
        }
        if(ans!=3 || pos!=1) {
            a[i].s3[1]=&#039;E&#039;;
            a[i].s3[2]=&#039;R&#039;;
            a[i].s3[3]=&#039;R&#039;;
        }
        //特判字符个数

        ans=0,pos=0,o=0,u=0;
        //归零

    }

    for(i=1; i<=n; i++) {
        if(a[i].s3[1]==&#039;E&#039;)continue;
        for(q=1; q<=n; q++) {
            if(a[q].s3[1]==&#039;E&#039;)continue;
            if(a[i].s2.size() != a[q].s2.size())continue;
            //特判，加快循环

            for(j=1; j<=a[i].s2.size(); j++) {
                if(a[i].s2[j] != a[q].s2[j]) {
                    f=false;
                    break;
                }
            }
            if(f==true) {
                if(a[i].s1[1]==&#039;S&#039; && a[q].s1[q]==&#039;C&#039;) {
                    a[i].s3[1]=&#039;O&#039;;
                    a[i].s3[2]=&#039;K&#039;;
                    a[q].s3[1]=(char)a[i].num;
                }

                if(a[i].s1[1]==&#039;C&#039; && a[q].s1[q]==&#039;S&#039;) {
                    a[q].s3[1]=&#039;O&#039;;
                    a[q].s3[2]=&#039;K&#039;;
                    a[i].s3[1]=(char)a[i].num;
                }
                //

            }
        }
    }
    for(i=1; i<=n; i++) {
        for(j=1; j<=strlen(a[i].s3); j++) {
            cout<<a[i].s3[j];
        }
        cout<<endl;
    }
    return 0;
}

[CSP-J 2022] 解密
RTE，要用二分算法，但不会，打算明天找个时间做一下
用的暴力解

#include<bits/stdc++.h>
using namespace std;
int n,e,d,f,k;

int main( ) {
    freopen("decode.in","r",stdin);
    freopen("decode.out","w",stdout);
    cin>>k;
    for(int i=1; i<=k; i++) {
        cin>>n>>d>>e;
        for(int j=1; j<=sqrt(n); j++) {
            if((j+n/j==n-e*d+2) && (n%j==0)) {
                cout<<j<<" "<<n/j<<"\n\n";
                f=1;
                break;
            }
        }
        if(!f) {
            cout<<"NO\n\n";
        } else {
            f=0;
        }
    }
    return 0;
}

题库水题复习以前的算法，不具体讨论