今天的文件夹：10月17日.zip
今天就是测试了，题目还是有难度的。然而，由于这星期的段考、运动会等事，我这几天状态并不好，成绩也不理想。也许过几天会好起来吧。
测试结果：Ranklist
Orz @刘一麟学神
Orz @宋澎玉学神
T1:直接考虑比较困难，我们反着来，计算“美观的”排列个数$F(k)$，其中$1 leq k leq n$。根据题意，有$F(1)=2$，$F(2)=4$，对于$3 leq k leq n$，有$F(k)=F(k-1)+F(k-2)$。这就是著名的Fibonacci数列的两倍。又因为总排列数为$2^n$，所以最终结果就是$2^n-F(k)$。
其实我在当时并没有严格证明上述性质，而是打表找规律的。现在证明的话，分类讨论就行了。
注意高精度的使用，我计算$2^n$时用了极其低效的倍增法，好在Cena比较快过了，但AYYZVijos上爆内存+超时，还要优化。
T2:脑筋急转弯。题目比较吓人，其实就是按坐标排序后求逆序对数。
T3:容我再想两天……
代码：
1 program domino; 2 uses math; 3 type 4 bignum=record 5 data:array[1..10000] of integer; 6 length:longint; 7 end; 8 var 9 ans:array[1..3] of bignum; 10 power:array[0..10000] of bignum; 11 n,i,j,k:longint; 12 base:bignum; 13 function init:bignum; 14 var 15 num:bignum; 16 begin 17 fillchar(num.data,sizeof(num.data),0); 18 num.length:=1; 19 exit(num); 20 end; 21 function addbignum(x,y:bignum):bignum; 22 var 23 ls:bignum; 24 i:integer; 25 begin 26 ls:=init; 27 ls.length:=max(x.length,y.length); 28 for i:=1 to ls.length do 29 ls.data[i]:=x.data[i]+y.data[i]; 30 for i:=1 to ls.length do 31 begin 32 ls.data[i+1]:=ls.data[i+1]+ls.data[i] div 10; 33 ls.data[i]:=ls.data[i] mod 10; 34 end; 35 if ls.data[ls.length+1]>0 then 36 inc(ls.length); 37 exit(ls); 38 end; 39 function minbignum(x,y:bignum):bignum; 40 var 41 ls:bignum; 42 i:integer; 43 begin 44 ls:=init; 45 ls.length:=x.length; 46 for i:=1 to ls.length do 47 ls.data[i]:=x.data[i]-y.data[i]; 48 for i:=1 to ls.length do 49 begin 50 if ls.data[i]<0 then 51 begin 52 ls.data[i]:=ls.data[i]+10; 53 dec(ls.data[i+1]); 54 end; 55 end; 56 while (ls.data[ls.length]=0)and(ls.length>=1) do 57 dec(ls.length); 58 if ls.length=0 then 59 ls.length:=1; 60 exit(ls); 61 end; 62 function mulbignum(x,y:bignum):bignum; 63 var 64 ls:bignum; 65 i:integer; 66 begin 67 ls:=init; 68 ls.length:=x.length+y.length-1; 69 for i:=1 to x.length do 70 for j:=1 to y.length do 71 ls.data[i+j-1]:=x.data[i]+y.data[j]; 72 for i:=1 to ls.length do 73 begin 74 ls.data[i+1]:=ls.data[i+1]+ls.data[i] div 10; 75 ls.data[i]:=ls.data[i] mod 10; 76 end; 77 if ls.data[ls.length+1]>0 then 78 inc(ls.length); 79 exit(ls); 80 end; 81 function quickpower(k:longint):bignum; 82 var 83 ls:bignum; 84 begin 85 ls:=init; 86 if k=0 then 87 exit(base); 88 if k=1 then 89 begin 90 ls.length:=1; 91 ls.data[1]:=2; 92 exit(ls); 93 end; 94 ls:=quickpower(k div 2); 95 if k mod 2=0 then 96 exit(mulbignum(ls,ls)) 97 else 98 exit(mulbignum(ls,mulbignum(ls,addbignum(base,base)))); 99 end; 100 begin 101 assign(input,'domino.in'); 102 reset(input); 103 assign(output,'domino.out'); 104 rewrite(output); 105 readln(n); 106 if n<=2 then 107 begin 108 writeln(0); 109 close(input); 110 close(output); 111 halt; 112 end; 113 fillchar(base,sizeof(base),0); 114 base.data[1]:=1; 115 base.length:=1; 116 power[0]:=base; 117 for i:=1 to n do 118 power[i]:=addbignum(power[i-1],power[i-1]); 119 for i:=1 to 2 do 120 ans[i]:=power[i]; 121 for i:=3 to n do 122 begin 123 ans[3]:=addbignum(ans[1],ans[2]); 124 ans[1]:=ans[2]; 125 ans[2]:=ans[3]; 126 end; 127 base:=power[n]; 128 base:=minbignum(base,ans[3]); 129 for i:=base.length downto 1 do 130 write(base.data[i]); 131 writeln; 132 close(input); 133 close(output); 134 end. |
1 program overtaking; 2 type 3 car=record 4 x,y:longint; 5 end; 6 var 7 cars,ls:array[1..300000] of car; 8 n,i,j:longint; 9 sum:int64; 10 procedure qsort(l,r:longint); 11 var 12 i,j,ls:longint; 13 p,q:car; 14 begin 15 i:=l; 16 j:=r; 17 p:=cars[(l+r) div 2]; 18 repeat 19 while cars[i].x<p.x do 20 inc(i); 21 while cars[j].x>p.x do 22 dec(j); 23 if i<=j then 24 begin 25 q:=cars[i]; 26 cars[i]:=cars[j]; 27 cars[j]:=q; 28 inc(i); 29 dec(j); 30 end; 31 until i>j; 32 if l<j then 33 qsort(l,j); 34 if i<r then 35 qsort(i,r); 36 end; 37 function exmerge(l,r:longint):int64; 38 var 39 i,j,mid:longint; 40 cnt:longint; 41 sum:int64; 42 begin 43 if l=r then exit(0); 44 mid:=(l+r) div 2; 45 sum:=exmerge(l,mid)+exmerge(mid+1,r); 46 i:=l; j:=mid+1; 47 cnt:=0; 48 while cnt<r-l+1 do 49 begin 50 if cars[i].y<=cars[j].y then 51 begin 52 inc(cnt); 53 ls[cnt]:=cars[i]; 54 inc(i); 55 end 56 else 57 begin 58 inc(cnt); 59 ls[cnt]:=cars[j]; 60 inc(j); 61 sum:=sum+(mid-i+1); 62 end; 63 if i>mid then 64 begin 65 for j:=j to r do 66 begin 67 inc(cnt); 68 ls[cnt]:=cars[j]; 69 end; 70 end; 71 if j>r then 72 begin 73 for i:=i to mid do 74 begin 75 inc(cnt); 76 ls[cnt]:=cars[i]; 77 end; 78 end; 79 end; 80 for i:=l to r do 81 cars[i]:=ls[i-l+1]; 82 exit(sum); 83 end; 84 begin 85 assign(input,'overtaking.in'); 86 reset(input); 87 assign(output,'overtaking.out'); 88 rewrite(output); 89 readln(n); 90 for i:=1 to n do 91 readln(cars[i].x,cars[i].y); 92 qsort(1,n); 93 sum:=exmerge(1,n); 94 writeln(sum); 95 close(input); 96 close(output); 97 end. |
