POJ-3468-A Simple Problem with Integers （线段树 区间求和）

A Simple Problem with Integers

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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

一天了。。。研究这么点东西。。。sad。。。

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>using namespace std;#define LL long longconst int inf = 111111;LL add[inf << 2];LL sum[inf << 2];LL anst;void pushup (int ii)//区间向上更新{    sum[ii] = sum[ii * 2] + sum[ii * 2 + 1];}void pushdown (int ii,int m)//区间下放{    if ( add[ii] )    {        add[ii * 2 ] += add[ii];        add[ii * 2 + 1] += add[ii];        sum[ii * 2] += add[ii] * (m-(m >> 1));        sum[ii * 2 + 1] +=  add[ii] * (m >> 1);//因为在建树的时候右边总是 m>>1    左边不确定有多少个         add[ii] = 0;    }}void build (int ii,int l,int r){    add[ii] = 0;    if ( l == r )    {        scanf ("%I64d",&sum[ii]);        return ;    }    int m = ( l + r ) >> 1;    build (ii * 2,l,m);    build (ii * 2 + 1,m + 1,r);    pushup (ii);}LL query (int ii,int l,int r,int a,int b){    if ( a <= l && r <= b)    {        return sum[ii];    }    pushdown (ii,r - l + 1);    int m = (l + r)>>1;     LL ans = 0;    if ( a <= m )        ans += query (ii * 2,l,m,a,b);    if ( b > m)        ans += query (ii * 2 + 1,m + 1,r,a,b);    return ans;}void updata (int ii,int l,int r,int a,int b,int z){    if ( a <= l && r <= b)//区间被完全覆盖    {        add[ii] += z;        sum[ii] +=  (LL) ((r - l + 1) * z);//当前节点的值 也就是此节点以下值得和 都加上z   r - l + 1就代表区间长度        return ;    }    pushdown (ii,r - l + 1);    int m = ( r + l )>>1;    if ( a <= m )        updata (ii * 2,l,m,a,b,z);    if ( b > m)        updata ( ii * 2 + 1,m + 1,r,a,b,z);    pushup (ii);}int main (){    int n,q;    scanf ("%d%d",&n,&q);    build (1,1,n);    while (q --)    {        char op[5];        int a,b,c;        scanf ("%s",op);        if ( op[0] == 'Q')        {            scanf ("%d%d",&a,&b);            anst = query (1,1,n,a,b);//从1号节点开始  树的区间从1到n  待查询区间 从a到b            printf ("%I64d\n",anst);        }        else        {            scanf ("%d%d%d",&a,&b,&c);            updata (1,1,n,a,b,c);//从1号节点开始 树的区间 待更新区间  更新值        }    }    return 0;}