poj3468 A Simple Problem with Integers 2011-12-20

Posted 2020-06-17

A Simple Problem with Integers

Time Limit: 5000MSMemory Limit: 131072K

Total Submissions: 26062Accepted: 7202

Case Time Limit: 2000MS

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

 

POJ Monthly--2007.11.25, Yang Yi

 

_________________________________________________

给一串数，两种操作：一是将一段区间的每个数都加上一个数。二是询问一段区间的和。

_________________________________________________

线段树，没有其他多余的操作，也是以前学非递归的线段树时做的。

_________________________________________________

 

  1 program Stone;
  2 
  3 const T2=1 shl 17-1;
  4 
  5 var i,j:longint;
  6 
  7     n,q,x,y,z:int64;
  8 
  9     ans:int64;
 10 
 11     a,sx:array[1..2*T2+2]of int64;
 12 
 13     b,sum:array[0..100000]of int64;
 14 
 15 procedure add(x,y,z:int64);                  //修改
 16 
 17 var s,i:int64;
 18 
 19  begin
 20 
 21   x:=x+T2-1;y:=y+T2+1;s:=1;                    
 22 
 23   while (x xor y)<>1 do                     //非递归的线段树，主要采取二进制的思想吧。
 24 
 25    begin
 26 
 27     if (x and 1)=0 then begin
 28 
 29                          inc(a[x+1],z);i:=x+1;
 30 
 31                          repeat
 32 
 33                            sx[i]:=sx[i]+z*s;i:=i div 2;
 34 
 35                          until i=1;
 36 
 37                         end;
 38 
 39     if (y and 1)=1 then begin
 40 
 41                          inc(a[y-1],z);i:=y-1;
 42 
 43                          repeat
 44 
 45                            sx[i]:=sx[i]+z*s;i:=i div 2;
 46 
 47                          until i=1;
 48 
 49                         end;
 50 
 51     x:=x div 2;y:=y div 2;s:=s*2;
 52 
 53    end;
 54 
 55  end;
 56 
 57 procedure que(x,y:int64);                    //询问
 58 
 59 var r,l,s:int64;
 60 
 61  begin
 62 
 63   ans:=sum[y-1]-sum[x-2];
 64 
 65   x:=x+T2-1;y:=y+T2+1;
 66 
 67   r:=0;l:=0;s:=1;
 68 
 69   while (x xor y)<>1 do
 70 
 71    begin
 72 
 73     if (x and 1)=0 then begin l:=l+s;ans:=ans+sx[x+1];end;
 74 
 75     if (y and 1)=1 then begin r:=r+s;ans:=ans+sx[y-1];end;
 76 
 77     x:=x div 2;y:=y div 2;s:=s*2;
 78 
 79     ans:=ans+l*a[x]+r*a[y];
 80 
 81    end;
 82 
 83    while y<>1 do
 84 
 85     begin
 86 
 87      y:=y div 2;ans:=ans+a[y]*(l+r);s:=s*2;
 88 
 89     end;
 90 
 91    writeln(ans);
 92 
 93  end;
 94 
 95 procedure init;
 96 
 97 var c,sp:char;
 98 
 99  begin
100 
101   readln(n,q);
102 
103   for i:=1 to n do
104 
105    begin
106 
107     read(b[i]);sum[i]:=sum[i-1]+b[i];         //初始化
108 
109    end;
110 
111    readln;
112 
113   for i:=1 to q do
114 
115    begin
116 
117     read(c,sp,x,y);
118 
119     inc(x);inc(y);  
120 
121     if c=‘C‘ then begin read(z);add(x,y,z);end;
122 
123     if c=‘Q‘ then que(x,y);
124 
125     readln;
126 
127    end;
128 
129  end;
130 
131 begin
132 
133  assign(input,‘pku3468.in‘);assign(output,‘pku3468.out‘);
134 
135  reset(input);rewrite(output);
136 
137   init;
138 
139  close(input);close(output);
140 
141 end.
142 
143  
144 
145