A. Pashmak and Garden http://codeforces.com/contest/459/problem/A
水题，没过，一字之差，细节决定成败
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 void out(int a,int b,int c,int d){ 5 printf("%d %d %d %d ",a,b,c,d); 6 } 7 int main(){ 8 int x1,y1,x2,y2; 9 while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){ 10 int sx=min(x1,x2); 11 int sy=min(y1,y2); 12 int bx=max(x1,x2); 13 int by=max(y1,y2); 14 int dx=bx-sx; 15 int dy=by-sy; 16 if(dx&&dy&&dx!=dy){ 17 puts("-1"); 18 continue; 19 } 20 if(dx&&dy){ 21 if((x1<x2&&y1<y2)||(x2<x1&&y2<y1)){ 22 out(sx,by,bx,sy); 23 continue; 24 } 25 out(sx,sy,bx,by); 26 continue; 27 } 28 if(dx){ 29 out(sx,sy+dx,bx,by+dx); 30 continue; 31 } 32 out(sx+dy,sy,bx+dy,by); 33 } 34 return 0; 35 }
B. Pashmak and Flowers http://codeforces.com/contest/459/problem/B
分类讨论
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 typedef __int64 LL; 5 const int M=200010; 6 int a[M]; 7 int main(){ 8 int n; 9 while(~scanf("%d",&n)){ 10 for(int i=0;i<n;i++){ 11 scanf("%d",&a[i]); 12 } 13 int big=a[0],sma=a[0]; 14 for(int i=1;i<n;i++){ 15 big=max(big,a[i]); 16 sma=min(sma,a[i]); 17 } 18 printf("%d ",big-sma); 19 if(big==sma){ 20 printf("%I64d ",(LL)n*(n-1)/2); 21 continue; 22 } 23 LL sumbig=0,sumsma=0; 24 for(int i=0;i<n;i++){ 25 if(a[i]==big) sumbig++; 26 else if(a[i]==sma) sumsma++; 27 } 28 printf("%I64d ",sumbig*sumsma); 29 } 30 return 0; 31 }
C. Pashmak and Buses
分类讨论，可构造就有多种解了，我是按照高精度加法从1111构造到kkkk，
1 #include<cstdio> 2 const int M=1024; 3 int a[M][M]; 4 int main(){ 5 int n,k,d; 6 while(~scanf("%d%d%d",&n,&k,&d)){ 7 int tmp=1; 8 bool flag=false; 9 for(int i=0;i<d;i++){ 10 tmp*=k; 11 if(tmp>=n){ 12 flag=true; 13 break; 14 } 15 } 16 if(!flag){ 17 puts("-1"); 18 continue; 19 } 20 for(int j=1;j<=d;j++){ 21 a[1][j]=1; 22 } 23 for(int i=2;i<=n;i++){ 24 for(int j=1;j<=d;j++){ 25 a[i][j]=a[i-1][j]; 26 } 27 a[i][d]++; 28 for(int j=d;j>=1;j--){ 29 if(a[i][j]>k){ 30 a[i][j]=1; 31 a[i][j-1]++; 32 } 33 } 34 } 35 for(int j=1;j<=d;j++){ 36 for(int i=1;i<=n;i++){ 37 printf("%d ",a[i][j]); 38 } 39 puts(""); 40 } 41 } 42 return 0; 43 }
D. Pashmak and Parmida's problem http://codeforces.com/contest/459/problem/D
朴素n^2,树状数组优化nlogn
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<map> 5 #define mt(a,b) memset(a,b,sizeof(a)) 6 using namespace std; 7 typedef __int64 LL; 8 const int M=1000010; 9 class One_Tree_Array { //一维树状数组 10 typedef int typev; 11 typev a[M]; 12 public: 13 void init() { 14 mt(a,0); 15 } 16 int lowb(int t) { 17 return t&(-t); 18 } 19 void add(int i,typev v) { 20 for(; i<M; a[i]+=v,i+=lowb(i)); 21 } 22 typev sum(int i) { 23 typev s=0; 24 for(; i>0; s+=a[i],i-=lowb(i)); 25 return s; 26 } 27 }gx; 28 int a[M],b[M],l[M],r[M]; 29 map<int,int> mp; 30 int main(){ 31 int n; 32 while(~scanf("%d",&n)){ 33 for(int i=0;i<n;i++){ 34 scanf("%d",&a[i]); 35 b[i]=a[i]; 36 } 37 sort(b,b+n); 38 int lb=unique(b,b+n)-b; 39 mp.clear(); 40 for(int i=0;i<lb;i++){ 41 mp[b[i]]=i; 42 } 43 for(int i=0;i<n;i++){ 44 b[i]=mp[a[i]]; 45 } 46 gx.init(); 47 mt(r,0); 48 for(int i=n-1;i>=0;i--){ 49 r[b[i]]++; 50 gx.add(r[b[i]],1); 51 } 52 mt(l,0); 53 LL ans=0; 54 for(int i=0;i<n;i++){ 55 l[b[i]]++; 56 gx.add(r[b[i]],-1); 57 r[b[i]]--; 58 ans+=gx.sum(l[b[i]]-1); 59 } 60 printf("%I64d ",ans); 61 } 62 return 0; 63 }
当然树状数组也可以掏出线段树替代之
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<map> 5 #define mt(a,b) memset(a,b,sizeof(a)) 6 #define lrrt int L,int R,int rt 7 #define iall 1,M,1 8 #define imid int mid=(L+R)>>1 9 #define lson L,mid,rt<<1 10 #define rson mid+1,R,rt<<1|1 11 using namespace std; 12 typedef __int64 LL; 13 const int M=1000010; 14 int tree[M<<2]; 15 void pushup(int rt){ 16 tree[rt]=tree[rt<<1]+tree[rt<<1|1]; 17 } 18 void update(int x,int val,lrrt){ 19 if(L==R){ 20 tree[rt]+=val; 21 return ; 22 } 23 imid; 24 if(mid>=x) update(x,val,lson); 25 else update(x,val,rson); 26 pushup(rt); 27 } 28 int query(int x,int y,lrrt){ 29 if(x<=L&&R<=y) return tree[rt]; 30 imid; 31 int ans=0; 32 if(mid>=x) ans+=query(x,y,lson); 33 if(mid<y) ans+=query(x,y,rson); 34 return ans; 35 } 36 int a[M],b[M],l[M],r[M]; 37 map<int,int> mp; 38 int main(){ 39 int n; 40 while(~scanf("%d",&n)){ 41 for(int i=0;i<n;i++){ 42 scanf("%d",&a[i]); 43 b[i]=a[i]; 44 } 45 sort(b,b+n); 46 int lb=unique(b,b+n)-b; 47 mp.clear(); 48 for(int i=0;i<lb;i++){ 49 mp[b[i]]=i; 50 } 51 for(int i=0;i<n;i++){ 52 b[i]=mp[a[i]]; 53 } 54 mt(tree,0); 55 mt(r,0); 56 for(int i=n-1;i>=0;i--){ 57 r[b[i]]++; 58 update(r[b[i]],1,iall); 59 } 60 mt(l,0); 61 LL ans=0; 62 for(int i=0;i<n;i++){ 63 l[b[i]]++; 64 update(r[b[i]],-1,iall); 65 r[b[i]]--; 66 if(l[b[i]]-1>=1) 67 ans+=query(1,l[b[i]]-1,iall); 68 } 69 printf("%I64d ",ans); 70 } 71 return 0; 72 }
E. Pashmak and Graph http://codeforces.com/contest/459/problem/E
找到单调性，然后dp
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define mt(a,b) memset(a,b,sizeof(a)) 5 using namespace std; 6 const int M=300010; 7 struct E{ 8 int u,v,w; 9 friend bool operator <(E a,E b){ 10 return a.w<b.w; 11 } 12 }e[M]; 13 int dpnow[M],dppre[M]; 14 int main(){ 15 int n,m; 16 while(~scanf("%d%d",&n,&m)){ 17 for(int i=0;i<m;i++){ 18 scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w); 19 } 20 sort(e,e+m); 21 mt(dppre,0); 22 for(int i=0,j;i<m;i++){ 23 for(j=i+1;j<m&&e[i].w==e[j].w;j++); 24 for(int k=i;k<j;k++){ 25 dpnow[e[k].u]=dppre[e[k].u]; 26 dpnow[e[k].v]=dppre[e[k].v]; 27 } 28 for(int k=i;k<j;k++){ 29 dpnow[e[k].v]=max(dpnow[e[k].v],dppre[e[k].u]+1); 30 } 31 for(int k=i;k<j;k++){ 32 dppre[e[k].u]=dpnow[e[k].u]; 33 dppre[e[k].v]=dpnow[e[k].v]; 34 } 35 i=j-1; 36 } 37 int ans=0; 38 for(int i=1;i<=n;i++){ 39 ans=max(ans,dppre[i]); 40 } 41 printf("%d ",ans); 42 } 43 return 0; 44 }
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