POJ 3468 A Simple Problem with Integers 线段树

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.

线段树区间修改

交了好多次才AC

#include <iostream>
#include<stdio.h>

using namespace std;
struct node{
long long  right,left,value;
long long lazy_add;
bool exist;
};
node tree[1600020];
long long number[100005];
long long found[100005];
void build_tree(long long i,long long le,long long ri)
{

    tree[i].left=le;
    tree[i].right=ri;
    tree[i].exist=true;
    tree[i].value=0;
      tree[i].lazy_add=0;
    if(le==ri)
    {
        found[le]=i;
        tree[i].value=number[le];
        return ;
    }
    build_tree(i<<1,le,(le+ri)>>1);
    build_tree((i<<1)+1,((le+ri)>>1)+1,ri);
}
void complete(long long N)
{
    for(long long i=4*N-1;i>=1;i=i-2)
    {
        if(tree[i].exist==true)
            tree[i>>1].value=tree[i].value+tree[i-1].value;

    }
}
long long m=0;
void push_down(long long i)
{
    tree[i].value+= tree[i].lazy_add;
    tree[i<<1].lazy_add+= (tree[i].lazy_add)*(tree[i<<1].right-tree[i<<1].left+1)/(tree[i].right-tree[i].left+1);
     tree[(i<<1)+1].lazy_add+= (tree[i].lazy_add)*(tree[(i<<1)+1].right-tree[(i<<1)+1].left+1)/(tree[i].right-tree[i].left+1);
     tree[i].lazy_add=0;
}
void push_up(long long i,long long ar)
{    long long p=ar;
    while(i)
    {
        i=i/2;
        tree[i].value+=p;
    }
}
void query(long long i,long long le,long long ri)
{
    if((tree[i].lazy_add!=0)&&!(tree[i].left==le&&tree[i].right==ri))
        push_down(i);
    if(tree[i].left==le&&tree[i].right==ri)
     {


        m=m+tree[i].value+tree[i].lazy_add;
        return ;
    }
    i=i<<1;
    if(le<=tree[i].right)
    {
        if(ri<=tree[i].right)
            query(i,le,ri);
        else
            query(i,le,tree[i].right);
    }
    i=i+1;
    if(ri>=tree[i].left)
    {
        if(le>=tree[i].left)
             query(i,le,ri);
        else
            query(i,tree[i].left,ri);
    }

}
long long z;
void lazy_change(long long i,long long le,long long ri,long long z)
{
    if(tree[i].left==le&&tree[i].right==ri)
    {
       tree[i].lazy_add+=z*(ri-le+1);
       long long ar=z*(ri-le+1);
        push_up(i,ar);
        return ;
    }
    i=i<<1;
    if(le<=tree[i].right)
    {
        if(ri<=tree[i].right)
            lazy_change(i,le,ri,z);
        else
            lazy_change(i,le,tree[i].right,z);
    }
    i=i+1;
    if(ri>=tree[i].left)
    {
        if(le>=tree[i].left)
            lazy_change(i,le,ri,z);
        else
            lazy_change(i,tree[i].left,ri,z);
    }
}


int main()
{
long long N,Q;
long long i;
char s[2];
while(scanf("%lld%lld",&N,&Q)!=EOF)
{





for(i=0;i<1600020;i++)
    tree[i].exist=false;
for(i=1;i<=N;i++)
{
    scanf("%lld",&number[i]);
}
long long x,y;
build_tree(1,1,N);
complete(N);

while(Q--)
{
    scanf("%s",&s);
    if(s[0]=='Q')
    {
        m=0;
        scanf("%lld%lld",&x,&y);
        query(1,x,y);
        printf("%lld\n",m);

    }

     if(s[0]=='C')
    {
scanf("%lld%lld%lld",&x,&y,&z);
long long yu=0;
 lazy_change(1,x,y,z);


    }

}
}


    return 0;
}