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
原文地址：https://www.cnblogs.com/HocRiser/p/11372250.html
时间： 2024-11-13 10:59:37