poj-3468 A Simple Problem with Integers(线段树成段更新)

/*
A Simple Problem with Integers
Time Limit: 5000MS        Memory Limit: 131072K
Total Submissions: 44944        Accepted: 13169
Case Time Limit: 2000MS
Description

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.
Source

POJ Monthly--2007.11.25, Yang Yi


题意：略

*/
/*
本题我做的时候出了点差错，后来看了大神的才知道
*/
#include <iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL long long
const int maxn = 111111;
LL add[maxn<<2];//不用长整型wa
LL sum[maxn<<2];
void PushUp(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)
{
    if (add[rt])
    {
        add[rt<<1]+= add[rt];//和乘法(赋相同的值)的区别，乘法是直接赋值
        add[rt<<1|1]+=add[rt];//和乘法(赋相同的值)的区别，乘法是直接赋值
        sum[rt<<1] +=(m-(m>>1))*add[rt];//长度和 (l+r)/2-l一样即(r-l+1)-(r-l+1)/2==(l+r)/2-l
        sum[rt<<1|1] += (m>>1)*add[rt];
        add[rt] = 0;
    }
}
void cre(int l,int r,int rt)
{
    add[rt]=0;
    if(l==r)
    {
        scanf("%lld",&sum[rt]);
        return;
    }

    int m= (l+r)>>1;
    cre(lson);
    cre(rson);
    PushUp(rt);
}
void upd(int L,int R,int c,int l,int r,int rt)
{
   // printf("前面  L = %d R = %d l = %d r = %d rt = %d\n",L,R,l,r,rt);
    if(L<=l&&r<=R)
    {
        add[rt]+=c;//和乘法(赋相同的值)的区别，乘法是直接赋值
        sum[rt]+=(LL)c*(r-l+1);

       // printf("L = %d R = %d l = %d r = %d rt = %d sum[%d] = %d\n",L,R,l,r,rt,rt,sum[rt]);
        return ;
    }
    PushDown(rt,r-l+1);
    int m =(l+r)>>1;
    if(L<=m)
        upd(L,R,c,lson);
    if(R>m)
        upd(L,R,c,rson);
    PushUp(rt);
}
LL qur(int L,int R,int l,int r,int rt)
{
   // printf("L = %d R = %d l = %d r = %d rt = %d\n",L,R,l,r,rt);
   // system("pause");
    if(L<=l&&r<=R)
    {
        return sum[rt];
    }
    PushDown(rt,r-l+1);
    int m =(l+r)>>1;
    LL ret=0;
    if(L<=m)
    {
        ret+=qur(L,R,lson);
        //printf("ret = %d\n",ret);
    }

    if(R>m)
    {

        ret+=qur(L,R,rson);
       // printf("ret = %d\n",ret);
    }


    return ret;
}
int main()
{
    int N,Q,a,b,c,i;
    char ch[2];
    while(~scanf("%d %d",&N,&Q))
    {
        cre(1,N,1);
        for(i=1; i<=Q; i++)
        {
            scanf("%s",ch);
            if(ch[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                printf("%lld\n",qur(a,b,1,N,1));
            }
            else if(ch[0]=='C')
            {
                scanf("%d%d%d",&a,&b,&c);
                upd(a,b,c,1,N,1);
            }
        }
    }
    return 0;
}