Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 1042 Solved: 381
每次x=1时,每行一个整数,表示这次旅行的开心度
4 1 100 5 5 5 1 1 2 2 1 2 1 1 2 2 2 3 1 1 4
101 11 11
对于100%的数据, n ≤ 100000,m≤200000 ,data[i]非负且小于10^9
题解:嗯哼!!!果然是加强版的数据,可惜还是一遍AC了(HansBug:^_^,咦?最近phile呢?),其实和“上帝造题的7分钟2”基本上一样,就是它的强化版,重点在于处理优化掉一些不必要的操作即可(比如当某区间内的数字全部<=1呵呵呵),具体不再赘述,详见我的代码模板:线段树5(传送门在此)
1 var
2 i,j,k,l,m,n:longint;
3 a,b:array[0..1000000] of int64;
4 function max(x,y:longint):longint;inline;
5 begin
6 if x>y then max:=x else max:=y;
7 end;
8 function min(x,y:longint):longint;inline;
9 begin
10 if x<y then min:=x else min:=y;
11 end;
12
13 procedure built(z,x,y:longint);inline;
14 begin
15 if (x=y) then
16 begin
17 read(a[z]);
18 if a[z]<=1 then b[z]:=1 else b[z]:=0;
19 end
20 else
21 begin
22 built(z*2,x,(x+y) div 2);
23 built(z*2+1,(x+y) div 2+1,y);
24 a[z]:=a[z*2]+a[z*2+1];
25 if (b[z*2]=1) and (b[z*2+1]=1) then b[z]:=1 else b[z]:=0;
26 end;
27 end;
28 function op(z,x,y,l,r:longint):int64;inline;
29 var a2,a3:int64;
30 begin
31 if l>r then exit(0);
32 if b[z]=1 then exit(0);
33 if (x=l) and (y=r) and (l=r) then
34 begin
35 a2:=a[z];
36 a[z]:=trunc(sqrt(a[z]));
37 if a[z]<=1 then b[z]:=1;
38 exit(a[z]-a2);
39 end;
40 a2:=op(z*2,x,(x+y) div 2,l,min(r,(x+y) div 2));
41 a3:=op(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r);
42 a[z]:=a[z]+a2+a3;
43 if (b[z*2]=1) AND (b[z*2+1]=1) then b[z]:=1;
44 exit(a2+a3);
45 end;
46 function cal(z,x,y,l,r:longint):int64;inline;
47 var a2,a3:int64;
48 begin
49 if l>r then exit(0);
50 if (x=l) and (y=r) then exit(a[z]);
51 a2:=cal(z*2,x,(x+y) div 2,l,min(r,(x+y) div 2));
52 a3:=cal(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r);
53 exit(a2+a3);
54 end;
55 begin
56 readln(n);
57 built(1,1,n);
58 readln;
59 readln(m);
60 for i:=1 to m do
61 begin
62 readln(j,k,l);
63 case j of
64 1:writeln(cal(1,1,n,k,l));
65 2:op(1,1,n,k,l);
66 end;
67 end;
68 readln;
69 end.
70