5.16pkusc模拟赛2

A.字符串显示数字a+b problem(5.16)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
char c[30][10][10];
char p[30][10][10];
int cnt,num,x[3],ret[100];
int judge(int m){
    int counter=0;
    for(int i=1;i<=7;i++)
        for(int j=1;j<=5;j++)
            if(c[m][i][j]=='x') counter++;
    if(counter==9) return -1;
    if(counter==7) return 1;
    if(counter==20) return 0;
    if(counter==14) return 4;
    if(counter==11) return 7;
    if(counter==23) return 8;
    if(counter==21&&c[m][2][5]=='x') return 9;
    if(counter==21) return 6;
    if(counter==19&&c[m][6][1]=='x') return 2;
    if(counter==19&&c[m][2][5]=='x') return 3;
    if(counter==19) return 5;
}
int main(){
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
strcpy(p[1][1],"....x");
strcpy(p[1][2],"....x");
strcpy(p[1][3],"....x");
strcpy(p[1][4],"....x");
strcpy(p[1][5],"....x");
strcpy(p[1][6],"....x");
strcpy(p[1][7],"....x");
strcpy(p[2][1],"xxxxx");
strcpy(p[2][2],"....x");
strcpy(p[2][3],"....x");
strcpy(p[2][4],"xxxxx");
strcpy(p[2][5],"x....");
strcpy(p[2][6],"x....");
strcpy(p[2][7],"xxxxx");
strcpy(p[3][1],"xxxxx");
strcpy(p[3][2],"....x");
strcpy(p[3][3],"....x");
strcpy(p[3][4],"xxxxx");
strcpy(p[3][5],"....x");
strcpy(p[3][6],"....x");
strcpy(p[3][7],"xxxxx");
strcpy(p[4][1],"x...x");
strcpy(p[4][2],"x...x");
strcpy(p[4][3],"x...x");
strcpy(p[4][4],"xxxxx");
strcpy(p[4][5],"....x");
strcpy(p[4][6],"....x");
strcpy(p[4][7],"....x");
strcpy(p[5][1],"xxxxx");
strcpy(p[5][2],"x....");
strcpy(p[5][3],"x....");
strcpy(p[5][4],"xxxxx");
strcpy(p[5][5],"....x");
strcpy(p[5][6],"....x");
strcpy(p[5][7],"xxxxx");
strcpy(p[6][1],"xxxxx");
strcpy(p[6][2],"x....");
strcpy(p[6][3],"x....");
strcpy(p[6][4],"xxxxx");
strcpy(p[6][5],"x...x");
strcpy(p[6][6],"x...x");
strcpy(p[6][7],"xxxxx");
strcpy(p[7][1],"xxxxx");
strcpy(p[7][2],"....x");
strcpy(p[7][3],"....x");
strcpy(p[7][4],"....x");
strcpy(p[7][5],"....x");
strcpy(p[7][6],"....x");
strcpy(p[7][7],"....x");
strcpy(p[8][1],"xxxxx");
strcpy(p[8][2],"x...x");
strcpy(p[8][3],"x...x");
strcpy(p[8][4],"xxxxx");
strcpy(p[8][5],"x...x");
strcpy(p[8][6],"x...x");
strcpy(p[8][7],"xxxxx");
strcpy(p[9][1],"xxxxx");
strcpy(p[9][2],"x...x");
strcpy(p[9][3],"x...x");
strcpy(p[9][4],"xxxxx");
strcpy(p[9][5],"....x");
strcpy(p[9][6],"....x");
strcpy(p[9][7],"xxxxx");
strcpy(p[0][1],"xxxxx");
strcpy(p[0][2],"x...x");
strcpy(p[0][3],"x...x");
strcpy(p[0][4],"x...x");
strcpy(p[0][5],"x...x");
strcpy(p[0][6],"x...x");
strcpy(p[0][7],"xxxxx");
    for(int i=1;i<=7;i++){
        char ch;
        cnt=1;num=0;
        while((ch=getchar())!='\n'&&ch!=EOF){
            if(cnt!=1&&num==0) ch=getchar();
            num++;
            c[cnt][i][num]=ch;
            if(num==5) num=0,cnt++;
        }
    }
    int d=1;
    for(int i=1;i<cnt;i++){
        if(judge(i)==-1) d=2;
        else x[d]=x[d]*10+judge(i);
    }
    int ans=x[1]+x[2];
    cnt=0;
    for(;ans;ans=ans/10) ret[++cnt]=ans%10;
    for(int i=1;i<=7;i++){
        for(int j=cnt;j>=1;j--){
            for(int k=0;k<5;k++)
                printf("%c",p[ret[j]][i][k]);
            if(j!=1) printf(".");
        }
        printf("\n");
    }
    return 0;
}

B.
C.
D.给定n，求最小长度的连续数字和=n。n<=10^9(5.16)
设l+(l+1)+…+r=n
(l+r)*(r-l+1)/2=n
(l+r)*(r-l+1)=2*n
l+r,r-l+1一奇一偶，暴力？

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
int T,t,a,b,ans,ansl,ansr;
int main(){
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    scanf("%d",&T);
    while(T--){
        ans=0;
        scanf("%d",&t);t*=2;a=1;b=t;
        while(!(b&1)) b/=2,a*=2;
        if(a<b) {
            ans=a;
            ansl=(b-a+1)/2;
            ansr=(b+a-1)/2;
        }else{
            ans=b;
            ansl=(a-b+1)/2;
            ansr=(b+a-1)/2;
        }
        if(ans==1) ans=inf;
        for(int i=2;i<=(int)sqrt(t);i++) 
            if(b%i==0) {
                if(ans<i) break;
                else {
                    ans=i;
                    a=i;b=t/a;
                    ansl=(b-a+1)/2;
                    ansr=(b+a-1)/2;
                }
            }
        if(ans!=inf) {
            printf("%d = ",t/2);
            for(int i=ansl;i<ansr;i++) printf("%d + ",i);printf("%d\n",ansr);
        }
        else printf("IMOPOSSIBLE\n");
    }
    return 0;
}

E.齿轮相切可逆向转动，已知1号轮子的转速为1，顺时针，求其他轮子的转动情况。(n<=1000)(5.16)
相邻轮子连边，遍历。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#define N 1005
#define M 1000005
using namespace std;
struct A {int x,y,r;}a[N];
struct E {int to,nxt;}edge[M*2];
int T,n;
int dis[N],idx[N],tot;
queue<int>q;
double dist(int x1,int y1,int x2,int y2){
    return (double)sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 
}
bool judge(int i,int j){
    return dist(a[i].x,a[i].y,a[j].x,a[j].y)==(double)(a[i].r+a[j].r);
}
void addedge(int from,int to){
//  printf("%d %d\n",from,to);
    edge[tot].to=to;edge[tot].nxt=idx[from];idx[from]=tot++;
}
void bfs(){
    q.push(1);dis[1]=1;
    while(!q.empty()){
        int x=q.front();q.pop();
        for(int t=idx[x];t;t=edge[t].nxt){
            E e=edge[t];
            if(!dis[e.to]){
                dis[e.to]=dis[x]+1;
                q.push(e.to);
            }
        }
    }
}
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
void getans(int x,int y){
    int d=gcd(x,y);
    x=x/d;y=y/d;
    if(y==1) printf("%d ",x);
    else printf("%d/%d ",x,y);
}
int main(){
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    scanf("%d",&T);
    while(T--){
        tot=1;
        memset(a,0,sizeof(A));
        memset(edge,0,sizeof(E));
        memset(idx,0,sizeof(idx));
        memset(dis,0,sizeof(dis));
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            int x,y,r;
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].r);
            for(int j=1;j<i;j++)
                if(judge(i,j)) addedge(i,j),addedge(j,i);
        }
        bfs();
        for(int i=1;i<=n;i++){
            if(dis[i]) getans(a[1].r,a[i].r);
            if(!dis[i]) printf("not moving\n");
            else if(dis[i]&1) printf("clockwise\n");
            else printf("counterclockwise\n");
        }
    }
    return 0;
}

F.求平面内矩形面积并。(n<=100)(5.17)
按照横坐标离散化。考虑从下至上扫描每条边，cnt[i]记录第i条水平线段覆盖情况。扫描矩形下边界对应横坐标(已离散化)+1，矩形上边界对应横坐标-1。每扫描完一个，统计新增面积。+1/-1可以线段树维护。但n<=100。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100005
using namespace std;
struct S{
    double l,r,h;
    int f;
    bool operator < (const S &t)const{
        return h<t.h;
    }
}ss[N*2];//水平线段扫描线
struct seg{
    int l,r,cnt;
    double len;
}t[N*4*2];
int n,m,num,tot;
double ans;
double pos[N*2];
void getlen(int k){
    if(t[k].cnt) t[k].len=pos[t[k].r+1]-pos[t[k].l];
    else if(t[k].l==t[k].r) t[k].len=0;
    else t[k].len=t[k<<1].len+t[k<<1|1].len;
}
int binary(double x,int l,int r){
    while(l<r){
        int mid=(l+r)/2;
        if(pos[mid]<x) l=mid+1;
        else r=mid;
    }
    return l;
}

void build(int k,int l,int r){
    t[k].l=l;t[k].r=r;t[k].len=0;t[k].cnt=0;
    if(l==r) return;
    int mid=(r+l)/2;
    build(k<<1,l,mid);build(k<<1|1,mid+1,r);
}
void upd(int k,int a,int b,int val){
    if(k>8*N) return;
    int l=t[k].l,r=t[k].r;
    if(b<l||r<a) return;
    if(a<=l&&r<=b) {
        t[k].cnt+=val;
        getlen(k);
        return;
    }
    upd(k<<1,a,b,val);upd(k<<1|1,a,b,val);
    getlen(k);
}
int main(){
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    while(scanf("%d",&n)&&n!=0){
        num=0;ans=0;m=0;tot++;
        memset(ss,0,sizeof(S));
        memset(t,0,sizeof(seg));
        memset(pos,0,sizeof(pos));
        for(int i=1;i<=n;i++){
            double x1,x2,y1,y2;
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            num++;ss[num].l=x1;ss[num].r=x2;ss[num].h=y1;ss[num].f=1;//下边
            num++;ss[num].l=x1;ss[num].r=x2;ss[num].h=y2;ss[num].f=-1;//上边
            pos[num-1]=x1;pos[num]=x2;
        }
        sort(ss+1,ss+1+num);
        sort(pos,pos+1+num);//离散化
        pos[0]=-1;
        for(int i=1;i<=num;i++) if(pos[i]!=pos[i-1]) pos[++m]=pos[i];//去重 线段树[1,m]
        build(1,1,m);
        for(int i=1;i<=num;i++){//扫描线 
            int l=binary(ss[i].l,1,m+1);
            int r=binary(ss[i].r,1,m+1)-1;
            upd(1,l,r,ss[i].f);
            ans+=(ss[i+1].h-ss[i].h)*t[1].len;
            //printf("%.2lf\n",ans);
        }
        printf("Test case #%d\nTotal explored area: %.2lf\n\n",tot,ans);
    }
    return 0;
}