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Title Description Description

A row of n squares, beginning with an integer in each lattice. Some questions and modifications are presented dynamically: the form of questioning is the summation of all elements in a particular sub-interval [a, b]; the modified rule is to specify a lattice x, plus or minus a specific value, a. You are now asked to make the right answer to each question. 1≤n<100000, the total number of questions and modifications m<10000. Enter a description input Description

The input file first behaves as an integer n, followed by n rows n integers, representing the original integers in the lattice. followed by a positive integer m, followed by M-line, the M-query, the first integer to indicate the query code, the number 1 indicates an increase, the following two numbers x and a for the value on position x to increase a, ask the code 2 for the interval sum, followed by two integers for A and B, representing the interval between [A, a] and Outputs description Output Description

A total of M-lines, each integer sample entered as Sample input

6

4

5

6

2

1

3

4

1 3 5

2 1 4

1 1 9

2 2 6 sample outputs sample output

22

22 data range and hints, size & Hint

1≤n≤100000, m≤10000.

Tree-like array:

1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int n,m,t[1000];
 6 void Add (int k,int z)
 7 {
 8     while (k<=n)
 9     {         t[k]+=z;         k+=k& (-K);
(int x)-{+-
int '     ans=0;     (x) (
ans+=t[x)         ;         x-=x& (-X);
*     *     return ans;
$
int main ()
{     cin>>n;     i=1;i<=n;i++ for (int) (int.)         x;         cin>>x;         Add (i,x);     cin>>m;     i=1;i<=m;i++ for (int),     {         a,b,c,d,e int;
Panax Notoginseng         cin>>a;         if (a==1) {             cin>>b>>c;             Add (b,c);         (a==2) (+) (
+             cin>>d>>e;             printf ("%d\n", find (E)-find (d-1));
(         +
}     ), and return 0;
50}

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Line segment Tree:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std;
 5 int n,m;
 6 struct node{7 int l,r,sum;
 8};
9 node tr[100010*4]; Ten void built (int l,int r,int k)//build one {tr[k].l=l;tr[k].r=r; if (l==r) {scanf ("%d", &tr[k].sum); return;
} mid= Int (l+r)/2;
Built (l,mid,k*2); built (mid+1,r,k*2+1);
Tr[k].sum=tr[k*2].sum+tr[k*2+1].sum;    + void Change (int k,int pos,int x), {$ int l=tr[k].l;int r=tr[k].r;
if (l==r) {tr[k].sum=tr[k].sum+x;return;} int mid= (L+R)/2;
if (pos<=mid) change (k*2,pos,x);
if (pos>mid) change (k*2+1,pos,x);
Tr[k].sum=tr[k*2].sum+tr[k*2+1].sum;      30} int que (int k,int l,int R) (int.) {ans=0, if (l==tr[k].l&&tr[k].r==r) {return tr[k].sum;}
int mid= (TR[K].L+TR[K].R)/2;
if (l<=mid) Ans+=que (K*2,l,min (R,mid));
if (r>mid) Ans+=que (K*2+1,max (mid+1,l), R);
return ans; 36 {37} int main ()     scanf ("%d", &n);
Built (1,n,1);
scanf ("%d", &m);         Max for (int i=1;i<=m;i++) $ {a,x,a,d,c int, cin>>a; if (a==1) 45             {cin>>x>>a; (1,x,a); (a==2) 50 {51
cin>>c>>d;
printf ("%d\n", Que (1,c,d));
+ 0};  56}

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