3212: Pku3468 A Simple Problem with Integers
3212: Pku3468 A Simple Problem with Integers
3212: Pku3468 A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1053 Solved: 468
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000. The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000. Each of the next Q lines represents an operation. "C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000. "Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
HINT
The sums may exceed the range of 32-bit integers.
Source
题解：其实就是个英文版的线段树模板题，连乘法覆盖啥的都没有，也就是说所有的修改操作压根就没有顺序之分，可以爱怎么瞎搞怎么瞎搞= =
PS：话说我的线段树居然全方位怒踩树状数组是什麽节奏啊啊啊QAQ（上面的是树状数组，下面的是线段树）
树状数组：
1 /**************************************************************
2 Problem: 3212
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:268 ms
7 Memory:15852 kb
8 ****************************************************************/
9
10 var
11 b,c:array[0..1000005] of int64;
12 i,j,k,l,m,n:longint;a1:int64;ch:char;
13 procedure addb(x:longint;y:int64);
14 begin
15 while x>0 do
16 begin
17 inc(b[x],y);
18 dec(x,x and (-x));
19 end;
20 end;
21 procedure addc(x:longint;y:int64);
22 var z:int64;
23 begin
24 z:=x;if x=0 then exit;
25 while x<=n do
26 begin
27 inc(c[x],z*y);
28 inc(x,x and (-x));
29 end;
30 end;
31 function sumb(x:longint):int64;
32 begin
33 sumb:=0;if x=0 then exit;
34 while x<=n do
35 begin
36 inc(sumb,b[x]);
37 inc(x,x and (-x));
38 end;
39 end;
40 function sumc(x:longint):int64;
41 begin
42 sumc:=0;
43 while x>0 do
44 begin
45 inc(sumc,c[x]);
46 dec(x,x and (-x));
47 end;
48 end;
49 function summ(x:longint):int64;
50 begin
51 exit(sumb(x)*int64(x)+sumc(x-1));
52 end;
53 procedure add(x,y:longint;z:int64);
54 begin
55 addb(y,z);addc(y,z);
56 addb(x-1,-z);addc(x-1,-z);
57 end;
58 function sum(x,y:longint):int64;
59 begin
60 exit(summ(y)-summ(x-1));
61 end;
62 begin
63 readln(n,m);
64 fillchar(b,sizeof(b),0);
65 fillchar(c,sizeof(c),0);
66 for i:=1 to n do
67 begin
68 read(a1);
69 add(i,i,a1);
70 end;readln;
71 for i:=1 to m do
72 begin
73 read(ch);
74 case upcase(ch) of
75 'Q':begin
76 readln(j,k);
77 writeln(sum(j,k));
78 end;
79 'C':begin
80 readln(j,k,a1);
81 add(j,k,a1);
82 end;
83 end;
84 end;
85 end.线段树：
1 /**************************************************************
2 Problem: 3212
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:84 ms
7 Memory:15852 kb
8 ****************************************************************/
9
10 var
11 i,j,k,l,m,n:longint;a1:int64;ch:char;
12 a,b:array[0..1000005] of int64;
13 function min(x,y:longint):longint;
14 begin
15 if x<y then min:=x else min:=y;
16 end;
17 function max(x,y:longint):longint;
18 begin
19 if x>y then max:=x else max:=y;
20 end;
21 procedure built(z,x,y:longint);
22 begin
23 if x=y then
24 read(a[z])
25 else begin
26 built(z*2,x,(x+y) div 2);
27 built(z*2+1,(x+y) div 2+1,y);
28 a[z]:=a[z*2]+a[z*2+1];
29 end;
30 b[z]:=0;
31 end;
32 procedure add(z,x,y,l,r:longint;t:int64);
33 begin
34 if l>r then exit;
35 if (x=l) and (y=r) then
36 begin
37 inc(b[z],t);
38 exit;
39 end;
40 inc(a[z],t*int64(r-l+1));
41 add(z*2,x,(x+y) div 2,l,min(r,(x+y) div 2),t);
42 add(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r,t);
43 end;
44 function sum(z,x,y,l,r:longint;t:int64):int64;
45 var a1,a2:int64;
46 begin
47 if l>r then exit(0);
48 inc(t,b[z]);
49 if (x=l) and (y=r) then exit(a[z]+t*int64(y-x+1));
50 a1:=sum(z*2,x,(x+y) div 2,l,min(r,(x+y) div 2),t);
51 a2:=sum(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r,t);
52 exit(a1+a2);
53 end;
54 begin
55 readln(n,m);built(1,1,n);readln;
56 for i:=1 to m do
57 begin
58 read(ch);
59 case upcase(ch) of
60 'Q':begin
61 readln(j,k);
62 writeln(sum(1,1,n,j,k,0));
63 end;
64 'C':begin
65 readln(j,k,a1);
66 add(1,1,n,j,k,a1);
67 end;
68 end;
69 end;
70 end.