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=;
 long long sum[maxn<<],add[maxn<<];
 void pushup(int rt)
 {
     sum[rt]=sum[rt<<]+sum[rt<<|];
 }
 void pushdown(int rt,int ln,int rn)
 {
     if(add[rt])
     {
         sum[rt<<]+=add[rt]*ln;
         sum[rt<<|]+=add[rt]*rn;
         add[rt<<]+=add[rt];
         add[rt<<|]+=add[rt];
         add[rt]=;
     }
 }
 void build(int l,int r,int rt)
 {
     if(r==l)
     {
         scanf("%lld",&sum[rt]);
         return;
     }
     int m=(l+r)>>;
     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+);
        add[rt]+=c;
        return;
    }
    int m=(l+r)>>;
    pushdown(rt,m-l+,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)>>;
     pushdown(rt,m-l+,r-m);
     long long ans=;
     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();
     int n,m,x,y,z,q;
     char h;
     while(~scanf("%d%d",&n,&q))
     {
     memset(sum,,sizeof sum);
     memset(add,,sizeof add);
     build(,n,);
     for(int i=;i<=q;i++)
     {
         getchar();
         scanf("%c",&h);
         if(h=='Q')
         {
             scanf("%d%d",&x,&y);
             cout<<query(x,y,,n,)<<endl;
         }
         else if(h=='C')
         {
             scanf("%d%d%d",&x,&y,&z);
             update(x,y,z,,n,);
         }
     }
     }
     return ;
 }

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！