poj 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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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 abc" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q ab" 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

Hint

The sums may exceed the range of 32-bit integers.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
typedef long long lls;
lls a[maxn];
lls sum[maxn<<2],add[maxn<<2],ll[maxn<<2],rr[maxn<<2];
inline void pushup(int i){
    sum[i]=sum[i<<1]+sum[i<<1|1];
}
inline void pushdown(lls i,lls m){
    if(add[i]){
        sum[i<<1]+=add[i]*(m-(m>>1));
        sum[i<<1|1]+=add[i]*(m>>1);
        add[i<<1]+=add[i];
        add[i<<1|1]+=add[i];
        add[i]=0;
    }
}
void build(lls l,lls r,lls i){
    ll[i]=l;
    rr[i]=r;
    add[i]=0;
    if(l==r){
        sum[i]=a[l];
        return;
    }
    lls m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    build(l,m,ls);
    build(m+1,r,rs);
    pushup(i);
}
void update(lls l,lls r,lls v,lls i){
    if(ll[i]>=l&&rr[i]<=r){
        add[i]+=v;
        sum[i]+=(lls)(rr[i]-ll[i]+1)*v;
        return ;
    }
    pushdown(i,rr[i]-ll[i]+1);
    lls m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    if(l<=m)update(l,r,v,ls);
    if(m<r)update(l,r,v,rs);
    pushup(i);
}
lls query(lls l,lls r,lls i){
    if(ll[i]>=l&&rr[i]<=r){
       return sum[i];
    }
    pushdown(i,rr[i]-ll[i]+1);
    lls m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    lls ans=0;
    if(l<=m)ans+=query(l,r,ls);
    if(m<r)ans+=query(l,r,rs);
    return ans;
}
int main()
{
    int n,qq;
    int u,v,c;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&qq)){
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        build(1,n,1);
        char q[2];
        while(qq--){
            scanf("%s",&q);
            if(q[0]=='Q'){
                scanf("%d%d",&u,&v);
                printf("%lld\n",query(u,v,1));
            }
            else {
                scanf("%d%d%d",&u,&v,&c);
                update(u,v,c,1);
            }
        }
    }
}