Tree Recovery

链接：https://www.nowcoder.com/acm/contest/77/H
来源：牛客网

Tree Recovery

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld

题目描述

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.

 

输入描述:

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.

输出描述:

You need to answer all Q commands in order. One answer in a line.

输入

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

输出

4
55
9
15

#include<bits/stdc++.h>  
using namespace std;  
#define lson rt<<1  
#define rson rt<<1|1  
#define ll long long  
const int maxn=1e5+7;  
ll a[maxn],add[maxn<<2];  
struct node  
{  
    int left,right,mid;  
    ll sum;  
}c[maxn<<2];  
void build(int l,int r,int rt)  
{  
    add[rt]=0;  
    c[rt].left=l;  
    c[rt].right=r;  
    c[rt].mid=(l+r)>>1;  
    if(l==r)  
    {  
        c[rt].sum=a[l];  
        return;  
    }  
    build(l,c[rt].mid,lson);  
    build(c[rt].mid+1,r,rson);  
    c[rt].sum=c[lson].sum+c[rson].sum;  
}  
void Push_down(int rt)  
{  
    if(add[rt]!=0)  
    {  
        add[lson]+=add[rt];  
        add[rson]+=add[rt];  
        //这里注意，add[lson],add[rson]此时不一定已经更新了，  
        //因此是在原来的基础上加，不是直接赋值  
        c[lson].sum+=(c[lson].right-c[lson].left+1)*add[rt];  
        c[rson].sum+=(c[rson].right-c[rson].left+1)*add[rt];  
        add[rt]=0;  
    }  
}  
void update(int l,int r,ll C,int rt)  
{  
    if(l<=c[rt].left&&r>=c[rt].right)  
    {  
        add[rt]+=C;  
        c[rt].sum+=(c[rt].right-c[rt].left+1)*C;  
        return;  
    }  
    Push_down(rt);  
    if(l<=c[rt].mid)update(l,r,C,lson);  
    if(r>c[rt].mid)update(l,r,C,rson);  
    c[rt].sum=c[lson].sum+c[rson].sum;  
}  
ll query(int l,int r,int rt)  
{  
    if(l<=c[rt].left&&r>=c[rt].right)  
        {  
            return c[rt].sum;  
        }  
    Push_down(rt);  
    int m=(c[rt].left+c[rt].right)>>1;  
    ll ans=0;  
    if(l<=m)ans+=query(l,r,lson);  
    if(r>m) ans+=query(l,r,rson);  
    return ans;  
}  
int main()  
{  
    ll n,Q,i,j;  
    scanf("%lld%lld",&n,&Q);  
    for(i=1;i<=n;i++)  
        scanf("%lld",&a[i]);  
    build(1,n,1);  
    while(Q--)  
    {  
        char t[5];int x,y;  
        scanf("%s%d%d",&t,&x,&y);  
        if(t[0]=='C')  
        {  
            ll z;scanf("%lld",&z);  
            update(x,y,z,1);  
        }  
        else  
        {  
             printf("%lld
",query(x,y,1));  
        }  
    }  
    return 0;  
}  

#include<stdio.h>
int main()
{
    int n,m,x,y,sum,i,z,a[100002];
    char f;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
        getchar();
        while(m--)
    {
        sum=0;
        scanf("%c",&f);
  
    if(f=='C')
    {scanf("%d%d%d",&x,&y,&z);
        for(i=x;i<=y;i++)
            a[i]+=z;
    }
    if(f=='Q')
    {
        scanf("%d%d",&x,&y);
        for(i=x;i<=y;i++)
            sum+=a[i];
        printf("%d
",sum);
    }
    getchar();
    }
    return 0;
}