POJ3468（线段树区间求和+区间查询）

https://vjudge.net/contest/66989#problem/C

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

//线段树区间和裸题
//#include<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<string.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=100005;
long long sum[maxn<<2],add[maxn<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int ln,int rn)
{
    if(add[rt])
    {
        sum[rt<<1]+=add[rt]*ln;
        sum[rt<<1|1]+=add[rt]*rn;
        add[rt<<1]+=add[rt];
        add[rt<<1|1]+=add[rt];
        add[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    if(r==l)
    {
        scanf("%lld",&sum[rt]);
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
   if(L<=l&&R>=r)
   {
       sum[rt]+=(long long)c*(r-l+1);
       add[rt]+=c;
       return;
   }
   int m=(l+r)>>1;
   pushdown(rt,m-l+1,r-m);
   if(L<=m)
    update(L,R,c,lson);
   if(R>m)
    update(L,R,c,rson);
    pushup(rt);
}
long long query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        return sum[rt];
    }
    int m=(r+l)>>1;
    pushdown(rt,m-l+1,r-m);
    long long ans=0;
    if(L<=m)
        ans+=query(L,R,lson);
    if(R>m)
        ans+=query(L,R,rson);
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m,x,y,z,q;
    char h;
    while(~scanf("%d%d",&n,&q))
    {
    memset(sum,0,sizeof sum);
    memset(add,0,sizeof add);
    build(1,n,1);
    for(int i=1;i<=q;i++)
    {
        getchar();
        scanf("%c",&h);
        if(h=='Q')
        {
            scanf("%d%d",&x,&y);
            cout<<query(x,y,1,n,1)<<endl;
        }
        else if(h=='C')
        {
            scanf("%d%d%d",&x,&y,&z);
            update(x,y,z,1,n,1);
        }
    }
    }
    return 0;
}

注意点：

是道裸题了

BUT 在输入数据的时候 ，scanf("%lld",&sum[rt]);一开始把它写成%d了，害我WA了好几发；

sum[rt]+=(long long)c*(r-l+1); 一开始没注意把数据转化为long long（＞人＜；）。

以后一定要注意呀！