20161005模拟

Posted 2020-08-12 神犇(shenben)

试题+评测数据下载：https://yunpan.cn/cvKJgPjvrIKAJ （提取码：2d8e）

分析：

T1 模拟计算几何问题 难点主要是圆的相交问题

T2 约瑟夫环问题，模拟即可

T3 稍微想想就能过，maxans:尽量多填数；minans：尽量少填数（满足主视图&左视图的情况下）

 

2bc*cosA=b^2+c^2-a^2

3

 

 

T1

AC代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double PI=3.14159265358979323846264; 
struct node{
    double x,y;
    double r;
}a[2];
inline double getdis(node b,node c){
    double xx=b.x-c.x;
    double yy=b.y-c.y;
    return sqrt(xx*xx+yy*yy);
}
void deal(node b,node c){
    double len=getdis(b,c);
    if(len<=fabs(b.r-c.r)){
        if(b.r<c.r) swap(b,c);
        double t1=PI*b.r*b.r;
        printf("%.3lf\\n",t1);
        return ;
    }
    double L=b.r+c.r;
    double t1=PI*b.r*b.r;
    double t2=PI*c.r*c.r;
    if(L<=len){
        printf("%.3lf\\n",t1+t2);
        return ;
    }
    double ang1=acos((b.r*b.r+len*len-c.r*c.r)/2.0/b.r/len);  
    double ang2=acos((c.r*c.r+len*len-b.r*b.r)/2.0/c.r/len);
    double ret=ang1*b.r*b.r+ang2*c.r*c.r-len*b.r*sin(ang1);   
    printf("%.3lf\\n",t1+t2-ret);
}
int main(){
    freopen("standing.in","r",stdin);
    freopen("standing.out","w",stdout);
    int T;scanf("%d",&T);
    while(T--){
        scanf("%lf%lf%lf%lf%lf%lf",&a[0].x,&a[0].y,&a[0].r,&a[1].x,&a[1].y,&a[1].r);
        deal(a[0],a[1]);
    }
    return 0;
}

 

T2

AC代码1：

#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e6+10;
int n,t;
bool vis[N];
void deal(int m){
    int tot=n;
    int p=0;
    for(int i=1;tot>1;i++){
        if(i>n) i=1;
        if(vis[i]) continue;
        if(++p==m) p=0,vis[i]=1,tot--;
    }
    for(int i=1;i<=n;i++) if(!vis[i]){printf("%d\\n",i);return ;}
}
int main(){
    freopen("resist.in","r",stdin);
    freopen("resist.out","w",stdout);
    scanf("%d%d",&n,&t);
    deal(t);
    return 0;
}

AC代码2：

#include<cstdio>
using namespace std;
int n,m,s;
int main(){
    freopen("resist.in","r",stdin);
    freopen("resist.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=2;i<=n;i++) s=(s+m)%i;
    printf("%d",s+1);
    return 0;
}

 

 

T3

AC代码1：

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e3+10;
int n,m,zhu[N],zuo[N],mp[N][N];
bool vis1[N],vis2[N];
int minans,maxans;
void get_min(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(!vis1[i]&&!vis2[j]&&zhu[i]==zuo[j]){
                vis1[i]=vis2[j]=1;
                mp[i][j]=zhu[i];
            }
        }
    }
    for(int i=1;i<=n;i++){
        if(!vis1[i]){
            for(int j=1;j<=m;j++){
                if(!mp[i][j]&&zhu[i]<=zuo[j]){
                    mp[i][j]=zhu[i];
                    if(zhu[i]==zuo[j]) vis2[j]=1;
                    break;
                }
            }
        }
    }
    for(int j=1;j<=m;j++){
        if(!vis2[j]){
            for(int i=1;i<=n;i++){
                if(!mp[i][j]&&zuo[j]<=zhu[i]){
                    mp[i][j]=zuo[j];break;
                }
            }
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            minans+=mp[i][j];
        }
    }
}
void get_max(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            maxans+=min(zhu[i],zuo[j]);
        }
    }
}
int main(){
    freopen("neighbor.in","r",stdin);
    freopen("neighbor.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",zhu+i);
    for(int i=1;i<=m;i++) scanf("%d",zuo+i);
    get_min();
    get_max();
    printf("%d %d",minans,maxans);
    return 0;
}

AC代码2：

#include<iostream>
#include<cstdio>
#include<cstring>
const int N=1e3+10;
using namespace std;
int n,m,zhu[N],zuo[N],maxans,minans;
int f[N],c[N];
int main(){
    freopen("neighbor.in","r",stdin);
    freopen("neighbor.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&zhu[i]);
    for(int i=1;i<=m;i++)scanf("%d",&zuo[i]);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            maxans+=min(zhu[i],zuo[j]);
    for(int i=1;i<=n;i++)f[zhu[i]]++;
    for(int i=1;i<=m;i++)c[zuo[i]]++;
    for(int i=0;i<=1000;i++) minans+=max(f[i],c[i])*i;
    printf("%d %d\\n",minans,maxans);
    return 0;    
}