2019年百度之星·程序设计大赛 初赛一 题解

Posted 2022-07-31 hocriser

1001 Polynomial:若第一个多项式的次数大于第二个，就是1/0，若小于就是0/1，若等于就是第一个多项式最高次项系数/第二个多项式最高次项系数。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 5 typedef long long ll;
 6 using namespace std;
 7 
 8 const int N=10010;
 9 int T,n,f[N],g[N];
10 
11 int main()
12     freopen("a.in","r",stdin);
13     freopen("a.out","w",stdout);
14     for (scanf("%d",&T); T--; )
15         scanf("%d",&n); int s1=n-1,s2=n-1;
16         rep(i,0,n) f[i]=g[i]=0;
17         rep(i,0,n-1) scanf("%d",&f[i]);
18         rep(i,0,n-1) scanf("%d",&g[i]);
19         while (!f[s1]) s1--;
20         while (!g[s2]) s2--;
21         if (s1>s2) puts("1/0"); continue; 
22         if (s1<s2) puts("0/1"); continue; 
23         int d=__gcd(f[s1],g[s2]); printf("%d/%d\n",f[s1]/d,g[s2]/d);
24     
25     return 0;
26 

1001

1002 Game:做法很多，下面是我的做法。

首先起点一定在某个区间端点上，枚举起点。然后每次需要做的就是，计算从目前位置到下一个区间最少需要多少步。这个直接就是当前位置到下一个区间的较近端点的距离/2上取整，唯一需要考虑的是最后一步是走一格还是两格。这时找到这两个位置分别最多到之后哪个区间为止都不需要移动，若不同则选择更靠后的那个。若相同则计算到那个区间的距离，取较小的那个。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 5 typedef long long ll;
 6 using namespace std;
 7 
 8 const int N=100010,inf=1e9;
 9 int T,n,a[N],l[N],r[N],L[N],R[N],L1[N],R1[N],L2[N],R2[N];
10 
11 int main()
12     freopen("b.in","r",stdin);
13     freopen("b.out","w",stdout);
14     for (scanf("%d",&T); T--; )
15         scanf("%d",&n); int tot=0,ans=inf;
16         rep(i,1,n) scanf("%d%d",&l[i],&r[i]),a[++tot]=l[i],a[++tot]=r[i],L1[i]=R1[i]=L2[i]=R2[i]=n+1;
17         rep(i,1,n-1)
18             rep(j,i+1,n)
19                 if (l[j]>l[i]) L[i]=1; L1[i]=j; break; 
20                 if (r[j]<l[i]) L[i]=0; L1[i]=j; break; 
21             
22             rep(j,i+1,n)
23                 if (r[j]<r[i]) R[i]=1; R1[i]=j; break; 
24                 if (l[j]>r[i]) R[i]=0; R1[i]=j; break; 
25             
26             rep(j,i+1,n) if (l[j]>l[i]+1 || r[j]<l[i]+1) L2[i]=j; break; 
27             rep(j,i+1,n) if (l[j]>r[i]-1 || r[j]<r[i]-1) R2[i]=j; break; 
28         
29         rep(i,1,tot)
30             int x=a[i],res=0;
31             rep(j,1,n)
32                 if (l[j]<=x && r[j]>=x) continue;
33                 if (l[j]>x)
34                     res+=(l[j]-x+1)/2;
35                     if ((l[j]-x)&1 && j<n && r[j]>l[j]) if (L2[j]>=L1[j] && L[j]) x=l[j]+1; else x=l[j];  else x=l[j];
36                 else
37                     res+=(x-r[j]+1)/2;
38                     if ((x-r[j])&1 && j<n && r[j]>l[j]) if (R2[j]>=R1[j] && R[j]) x=r[j]-1; else x=r[j];  else x=r[j];
39                 
40             
41             ans=min(ans,res);
42         
43         printf("%d\n",ans);
44     
45     return 0;
46 

1002

1003 Mindis:离散化连边后跑最短路。

 1 #include<queue>
 2 #include<vector>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 7 using namespace std;
 8 
 9 const int N=200010,M=410;
10 double d[N];
11 bool b[N];
12 int T,n,xs,ys,xe,ye,xc,yc,xa[N],xb[N],ya[N],yb[N],xx[N],yy[N],ax[M][M],ay[M][M];
13 struct P int x; double d; ;
14 bool operator <(const P &a,const P &b) return a.d>b.d; 
15 vector<P>a[N];
16 priority_queue<P>Q;
17 
18 int main()
19     freopen("d.in","r",stdin);
20     freopen("d.out","w",stdout);
21     for (scanf("%d",&T); T--; )
22         scanf("%d",&n); xc=yc=0;
23         memset(ax,0,sizeof(ax)); memset(ay,0,sizeof(ay)); memset(b,0,sizeof(b));
24         rep(i,1,n)
25             scanf("%d%d%d%d",&xa[i],&ya[i],&xb[i],&yb[i]);
26             xx[++xc]=xa[i]; xx[++xc]=xb[i]; yy[++yc]=ya[i]; yy[++yc]=yb[i];
27         
28         scanf("%d%d%d%d",&xs,&ys,&xe,&ye);
29         xx[++xc]=xs; xx[++xc]=xe; yy[++yc]=ys; yy[++yc]=ye;
30         sort(xx+1,xx+xc+1); sort(yy+1,yy+yc+1);
31         xc=unique(xx,xx+xc+1)-xx-1; yc=unique(yy,yy+yc+1)-yy-1;
32         rep(i,1,xc*yc) a[i].clear();
33         rep(i,1,n)
34             xa[i]=lower_bound(xx+1,xx+xc+1,xa[i])-xx;
35             xb[i]=lower_bound(xx+1,xx+xc+1,xb[i])-xx;
36             ya[i]=lower_bound(yy+1,yy+yc+1,ya[i])-yy;
37             yb[i]=lower_bound(yy+1,yy+yc+1,yb[i])-yy;
38         
39         xs=lower_bound(xx+1,xx+xc+1,xs)-xx; xe=lower_bound(xx+1,xx+xc+1,xe)-xx;
40         ys=lower_bound(yy+1,yy+yc+1,ys)-yy; ye=lower_bound(yy+1,yy+yc+1,ye)-yy;
41         rep(i,1,n) rep(j,xa[i],xb[i]-1) rep(k,ya[i],yb[i]) ax[j][k]++;
42         rep(i,1,n) rep(j,xa[i],xb[i]) rep(k,ya[i],yb[i]-1) ay[j][k]++;
43         rep(i,1,xc) rep(j,1,yc-1)
44             a[(i-1)*yc+j].push_back((P)(i-1)*yc+j+1,(yy[j+1]-yy[j])/(ay[i][j]+1.));
45             a[(i-1)*yc+j+1].push_back((P)(i-1)*yc+j,(yy[j+1]-yy[j])/(ay[i][j]+1.));
46         
47         rep(i,1,xc-1) rep(j,1,yc)
48             a[(i-1)*yc+j].push_back((P)i*yc+j,(xx[i+1]-xx[i])/(ax[i][j]+1.));
49             a[i*yc+j].push_back((P)(i-1)*yc+j,(xx[i+1]-xx[i])/(ax[i][j]+1.));
50         
51         rep(i,1,xc*yc) d[i]=1e30;
52         d[(xs-1)*yc+ys]=0; Q.push((P)(xs-1)*yc+ys,d[(xs-1)*yc+ys]);
53         while (Q.size())
54             int x=Q.top().x; Q.pop();
55             if (b[x]) continue; b[x]=1;
56             for (int i=0; i<(int)a[x].size(); i++)
57                 P e=a[x][i];
58                 if (d[e.x]>d[x]+e.d) d[e.x]=d[x]+e.d,Q.push((P)e.x,d[e.x]);
59             
60         
61         printf("%.5lf\n",d[(xe-1)*yc+ye]);
62     
63     return 0;
64 

1003

1005 Seq:打表找结论，具体见代码。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

int T;
ll n;

int main()
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
    for (scanf("%d",&T); T--; )
        scanf("%lld",&n);
        if (n%6==0 || n%6==2) printf("%lld\n",n/2);
        if (n%6==3 || n%6==5) printf("%lld\n",n/6);
        if (n%6==4) printf("%lld\n",n-1);
        if (n%6==1) printf("%lld\n",n/2+n/6+1);
    
    return 0;

1005