poj3468（一维）（区间查询，区间修改）

A Simple Problem with Integers

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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。

 

还是简单的线段树。

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 const int MAXN=;
 char s[],c;
 long long tree[MAXN*]={},add[MAXN*]={};
 int t,num,x,n,y,m,z,a[MAXN]={};
 using namespace std;
 void build(int l,int r,int p){
     if (l==r) tree[p]=a[l];
     else {
         int mid=(l+r)>>;
         build(l,mid,p*);
         build(mid+,r,p*+);
         tree[p]=tree[p*]+tree[p*+];
     }
 }
 long long query(int l,int r,int p,int x,int y){
     if (l==x&&r==y) return tree[p]+add[p]*(r-l+);
     else {
         add[p*+]+=add[p];
         add[p*]+=add[p];
         tree[p]+=add[p]*(r-l+);
         add[p]=;
         int mid=(l+r)>>;
         if (y<=mid) return query(l,mid,p*,x,y);
         else if(x>=mid+) return query(mid+,r,p*+,x,y);
              else return query(l,mid,p*,x,mid)+query(mid+,r,p*+,mid+,y);
     }
 }
 void change(int l,int r,int p,int x,int y,int zhi)
 {
     if (l==x&&r==y) add[p]+=zhi;
     else
     {
         tree[p]+=zhi*(y-x+);
         int mid=(l+r)>>;
         if (y<=mid) change(l,mid,p*,x,y,zhi);
         else if (x>mid) change(mid+,r,p*+,x,y,zhi);
         else
         {
             change(l,mid,p*,x,mid,zhi);
             change(mid+,r,p*+,mid+,y,zhi);
         }
     }
 }
 int main(){
     while (~scanf("%d%d",&n,&m)){
         for (int i=;i<=n;i++)
             scanf("%d",&a[i]);
         build(,n,);
         for (int i=;i<=m;i++){
             scanf("%s%d%d",&s,&x,&y);
             if (s[]=='Q') printf("%lld\n",query(,n,,x,y));
             else {
                 scanf("%d",&z);
                 change(,n,,x,y,z);
             }
         }
     }
 }

线段树虽然简单，但是树状数组就没有这么简单了，差分思想。

转换应该在树状数组blog上已经讲过了，会在blog上持续更新，最新的理解。

 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define N 100007
 using namespace std;

 int n,m;
 long long c[N],d[N];
 char ch[];

 int lowbit(int x)
 {
     return x&(-x);
 }
 void change(int x,int y,int z)
 {
     for (int i=x;i<=n;i+=lowbit(i))
         c[i]+=z;
     for (int i=y+;i<=n;i+=lowbit(i))
         c[i]-=z;

     for (int i=x;i<=n;i+=lowbit(i))
         d[i]+=(long long)z*x;
     for (int i=y+;i<=n;i+=lowbit(i))
         d[i]-=(long long)z*(y+);
 }
 long long query(int x)
 {
     long long res=,ans=;
     for (int i=x;i>=;i-=lowbit(i))
         res+=c[i];
     ans+=res*(x+);
     res=;
     for (int i=x;i>=;i-=lowbit(i))
         res+=d[i];
     ans-=res;
     return ans;
 }
 int main()
 {
     scanf("%d%d",&n,&m);
     int last=,x,y,z;
     for (int i=;i<=n;i++)
     {
         scanf("%d",&x);
         y=x-last;
         change(i,n,y);
         last=x;
     }
     for (int i=;i<=m;i++)
     {
         scanf("%s",ch);
         if (ch[]=='Q')
         {
             scanf("%d%d",&x,&y);
             printf("%lld\n",query(y)-query(x-));
         }
         else
         {
             scanf("%d%d%d",&x,&y,&z);
             change(x,y,z);
         }
     }
 }