线段树区间更新的两种方法（本质一样）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 126758		Accepted: 39405
Case Time Limit: 2000MS

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 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+88;
long long sum[N<<2],add[N<<2];
int n,q;
char op[5];
void up(int x,int l=0,int r=0){
    sum[x]=sum[x<<1]+add[x<<1]*1LL*((r-l)/2+1)+add[x<<1|1]*1LL*((r-l+1)/2)+sum[x<<1|1];
}
void build(int l,int r,int x){
    int m=(l+r)>>1;
    if(l==r) {
        scanf("%lld",&sum[x]);
        return;
    }
    build(l,m,x<<1);
    build(m+1,r,x<<1|1);
    up(x);
}
void update(int l,int r,int L,int R,int x,long long v){
    int m=(l+r)>>1;
    if(L<=l&&R>=r) {
        add[x]+=v;return;
    }
    if(L<=m) update(l,m,L,R,x<<1,v);
    if(R>m) update(m+1,r,L,R,x<<1|1,v);
    up(x,l,r);
}
long long query(int l,int r,int L,int R,int x,long long t){
    long long ret=0;
    t+=add[x];
    if(L<=l&&R>=r) return 1LL*t*(r-l+1)+sum[x];
    int m=(l+r)>>1;
    if(L<=m) ret+=query(l,m,L,R,x<<1,t);
    if(R>m) ret+=query(m+1,r,L,R,x<<1|1,t);
    return ret;
}
int main(){
    scanf("%d%d",&n,&q);
    build(1,n,1);
    while(q--){
        int x,y,c;
        scanf("%s",op);
        if(op[0]=='Q') {
            scanf("%d%d",&x,&y);
            printf("%lld
",query(1,n,x,y,1,0LL));
        }
        else {
            scanf("%d%d%d",&x,&y,&c);
            update(1,n,x,y,1,1LL*c);
        }
    }
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+88;
long long sum[N<<2],add[N<<2];
int n,q;
char op[5];
void up(int x){
    sum[x]=sum[x<<1]+sum[x<<1|1];
}
void pd(int x,int length){
    if(add[x]) {
        add[x<<1]+=add[x],add[x<<1|1]+=add[x];
        sum[x<<1]+=1LL*add[x]*(length-(length>>1));
        sum[x<<1|1]+=1LL*add[x]*(length>>1);
        add[x]=0;
    }
}
void build(int l,int r,int x){
    int m=(l+r)>>1;
    if(l==r) {
        scanf("%lld",&sum[x]);
        return;
    }
    build(l,m,x<<1);
    build(m+1,r,x<<1|1);
    up(x);
}
void update(int l,int r,int L,int R,int x,long long v){
    int m=(l+r)>>1;
    if(L<=l&&R>=r) {
        sum[x]+=1LL*(r-l+1)*v;
        add[x]+=v;
        return;
    }
    pd(x,r-l+1);
    if(L<=m) update(l,m,L,R,x<<1,v);
    if(R>m) update(m+1,r,L,R,x<<1|1,v);
    up(x);
}
long long query(int l,int r,int L,int R,int x){
    long long ret=0;
    if(L<=l&&R>=r) return sum[x];
    pd(x,r-l+1);
    int m=(l+r)>>1;
    if(L<=m) ret+=query(l,m,L,R,x<<1);
    if(R>m) ret+=query(m+1,r,L,R,x<<1|1);
    return ret;
}
int main(){
    scanf("%d%d",&n,&q);
    build(1,n,1);
    while(q--){
        int x,y,c;
        scanf("%s",op);
        if(op[0]=='Q') {
            scanf("%d%d",&x,&y);
            printf("%lld
",query(1,n,x,y,1));
        }
        else {
            scanf("%d%d%d",&x,&y,&c);
            update(1,n,x,y,1,1LL*c);
        }
    }
}