poj3468 A Simple Problem with Integers 2011-12-20

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

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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。

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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。

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  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