2019 年华东师范大学机试

 没有经过数据的检验，如果有错误欢迎指正

找规律后可以找到这个。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const double eps = 1e-6;
const ll N = 1e5+7;
int main(){
    ll a,g,c,k; cin>>a>>g>>c>>k;
    if(k&1){
        cout<<g-a<<endl;
    }else{
        cout<<a-g<<endl;
    }
    return 0;
}

 

固定任意一条边（我这里用了最大 然后二分角度）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll N = 1e5+7;
vector<int> v;
double len(double x1,double y1,double x2,double y2){
    double x=(x2-x1)*(x2-x1);
    double y=(y2-y1)*(y2-y1);
    return sqrt(x+y);
}
int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
int main(){
    int x;
    for(int i=0;i<3;i++){
        cin>>x;
        v.push_back(x);
    }
    sort(v.begin(),v.end());
    double l,r;
    l=0; r=pi/2;
    double ansx,ansy;
    //cout<<asin(3.0/5)<<endl;
    while(1){
        double mid=(l+r)/2;
        double xx=v[1]*cos(mid);
        double yy=v[1]*sin(mid);
        //cout<<mid<<" "<<len(xx,yy,v[2],0)<<" "<<v[0]<<endl;
        if(sgn(len(xx,yy,v[2],0)-v[0])==0){
            ansx=xx;
            ansy=yy;
            //cout<<ansx<<" "<<ansy<<endl;
            break;
        }else{
            if(sgn(len(xx,yy,v[2],0)-v[0])==1){
                r=mid;
            }else{
                l=mid;
            }
        }
    }
    cout<<"0 0"<<endl;
    cout<<v[2]<<" 0"<<endl;
    //cout<<ansx<<endl;
    printf("%.1f %.1f\n",ansx,ansy);
    //cout<<ansx<<" "<<ansy<<endl;
    return 0;
}

 

模拟即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const double eps = 1e-6;
const ll N = 1e5+7;
int ans[107];
int main(){
    int k; string s;
    cin>>k>>s;
    for(int i=0;i<s.length();i++){
        ans[i%k]+=s[i]-'0';
    }
    int c=0;
//    for(int i=0;i<k;i++){
//        cout<<ans[i];
//    }
    while(1){
        for(int i=k-1;i>=0;i--){
            int tmp=(ans[i]+c)%2;
            c=(ans[i]+c)/2;
            ans[i]=tmp;
        }
        if(c==0) break;
    }
    for(int i=0;i<k;i++){
        cout<<(!ans[i]);
    }
    cout<<endl;
    return 0;
}
//4
//101101100111
//0110

//16
//1111111111111111
//0000000000000000

//16
//10100000110000111110100101001000
//0111010111110011

 

待做