POJ 3468A Simple Problem with Integers(线段树区间更新)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 112228		Accepted: 34905
Case Time Limit: 2000MS

Description

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


题意就是如题，对线段树的区间进行更新，可增可减可修改，并查询区间和，此题就是增，具体实现看代码吧.
AC代码：

Source Code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define maxsize 100005
#define LL long long
LL num[maxsize];
struct node
{
    LL l,r;
    LL maxn,add;
    int mid()
    {
        return (l+r)/2;
    }
} tree[maxsize<<2];
void Build_Tree(LL root,LL l,LL r)
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].add=0;
    if(l==r)
    {
        tree[root].maxn=num[l];
        return ;
    }
    Build_Tree(root*2,l,tree[root].mid());
    Build_Tree(root*2+1,tree[root].mid()+1,r);
    tree[root].maxn=tree[root*2].maxn+tree[root*2+1].maxn;
}
void Update(LL root,LL l,LL r,LL czp)
{
    if(tree[root].l==l&&tree[root].r==r)
    {
        tree[root].add+=czp;
        return ;
    }
    tree[root].maxn+=((r-l+1)*czp);
    if(tree[root].mid()>=r) Update(root<<1,l,r,czp);
    else if(tree[root].mid()<l)     Update(root<<1|1,l,r,czp);
    else
    {
        Update(root<<1,l,tree[root].mid(),czp);
        Update(root<<1|1,tree[root].mid()+1,r,czp);
    }
}
long long Query(LL root,LL l,LL r)
{
    if(tree[root].l==l&&tree[root].r==r)
    {
        return tree[root].maxn+(r-l+1)*tree[root].add;
    }
    tree[root].maxn+=(tree[root].r-tree[root].l+1)*tree[root].add;
    Update(root<<1,tree[root].l,tree[root].mid(),tree[root].add);
    Update(root<<1|1,tree[root].mid()+1,tree[root].r,tree[root].add);
    tree[root].add=0;
    if(tree[root].mid()>=r) return Query(root*2,l,r);
    else if(tree[root].mid()<l) return Query(root*2+1,l,r);
    else
    {
        LL Lans=Query(root*2,l,tree[root].mid());
        LL Rans=Query(root*2+1,tree[root].mid()+1,r);
        return Lans+Rans;
    }
}
int main()
{
    /*ios::sync_with_stdio(false);*/
    char str[5];
    LL m,n,a,b,c;
    while(scanf("%lld%lld",&m,&n)!=EOF)
    {
        for(LL i=1; i<=m; i++)    scanf("%I64d",&num[i]);
        Build_Tree(1,1,m);
        for(LL i=1; i<=n; i++)
        {
            scanf("%s%I64d%I64d",str,&a,&b);
            if(str[0]=='Q')    printf("%I64d\n",Query(1,a,b));
            else    if(str[0]=='C')
            {
                scanf("%I64d",&c);
                Update(1,a,b,c);
            }
        }
    }
    return 0;
}