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poj 3468 A Simple Problem with Integers 线段树区间更新

2024-07-25 08:57:48   226人阅读

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A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 63565 Accepted: 19546
Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , AN. 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, ... , 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.

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.

给出n个数q次操作

C代表把a到b间的数分别加c

Q要求输出和

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=111111;
long long sum[MAXN<<2];
long long lazy[MAXN<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
    if(lazy[rt])
    {
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        sum[rt<<1]+=lazy[rt]*(m-(m>>1));
        sum[rt<<1|1]+=lazy[rt]*(m>>1);
        lazy[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    lazy[rt]=0;
    if(l==r)
    {
        scanf("%lld",&sum[rt]);
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(l>=L&R>=r)
    {
        lazy[rt]+=c;
        sum[rt]+=(long long)c*(r-l+1);
        return ;
    }
    pushdown(rt,r-l+1);
    int mid=(l+r)>>1;
    if(L<=mid)
        update(L,R,c,l,mid,rt<<1);
    if(R>mid)
        update(L,R,c,mid+1,r,rt<<1|1);
    pushup(rt);
}
long long query(int L,int R,int l,int r,int rt)
{
    if(l>=L&&R>=r)
    {
        return sum[rt];
    }
    pushdown(rt,r-l+1);
    int mid=(l+r)>>1;
    long long cnt=0;
    if(L<=mid)
        cnt=query(L,R,l,mid,rt<<1);
    if(R>mid)
        cnt+=query(L,R,mid+1,r,rt<<1|1);
    return cnt;
}
int main()
{
    int n,q;
    char op[2];
    int a,b,c;
    while(scanf("%d %d",&n,&q)!=EOF)
    {
        build(1,n,1);
        //printf("%d\n",sum[1]);
        while(q--)
        {
            scanf("%s",op);
            if(op[0]=='Q')
            {
                scanf("%d %d",&a,&b);
                printf("%lld\n",query(a,b,1,n,1));
            }
            else
            {
                scanf("%d %d %d",&a,&b,&c);
                update(a,b,c,1,n,1);
            }
        }
    }
    return 0;
}


poj 3468 A Simple Problem with Integers 线段树区间更新


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