POJ 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 110999		Accepted: 34570
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

4
55
9
15

线段树区间更新，区间求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long LL;
const int mod=100006;
LL n,m,x,y,z;
LL vis[mod<<2];
LL sum[mod<<2];
void pushup(LL node)
{
    sum[node]=sum[node<<1]+sum[node<<1|1];
}
void pushdown(LL node,LL m)
{
    if(vis[node])
    {
        vis[node<<1]+=vis[node];
        vis[node<<1|1]+=vis[node];
        sum[node<<1]+=vis[node]*(m-(m>>1));
        sum[node<<1|1]+=vis[node]*(m>>1);
        vis[node]=0;
    }
}
void build(LL l,LL r,LL node)
{
    vis[node]=0;
    if(l==r)
    {
        scanf("%lld",&sum[node]);
        return;
    }
    LL mid=(l+r)>>1;
    build(l,mid,node<<1);
    build(mid+1,r,node<<1|1);
    pushup(node);
}
void update(LL ll,LL rr,LL val,LL l,LL r,LL node)
{
    if(ll<=l && rr>=r)
    {
        vis[node]+=val;
        sum[node]+=val*(r-l+1);
        return ;
    }
    pushdown(node,r-l+1);
    LL mid=(l+r)>>1;
    if(ll<=mid) update(ll,rr,val,l,mid,node<<1);
    if(rr>mid) update(ll,rr,val,mid+1,r,node<<1|1);
    pushup(node);
}
LL query(LL ll,LL rr,LL l,LL r,LL node)
{
    LL ans=0;
    if(ll<=l && rr>=r)
        return sum[node];
    pushdown(node,r-l+1);
    LL mid=(l+r)>>1;
    if(ll<=mid) ans+=query(ll,rr,l,mid,node<<1);
    if(rr>mid) ans+=query(ll,rr,mid+1,r,node<<1|1);
    return ans;
}
int main()
{
    scanf("%lld%d",&n,&m);
    build(1,n,1);
    while(m--)
    {
        char a[5];
        scanf("%s",a);
        if(a[0]=='Q')
        {
            scanf("%lld%lld",&x,&y);
            printf("%lld\n",query(x,y,1,n,1));
        }
        else
        {
            scanf("%lld%lld%lld",&x,&y,&z);
            update(x,y,z,1,n,1);
        }
    }
    return 0;
}

 

 非递归实现

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const int maxn=1e5+6;
ll sum[maxn<<1];
ll add[maxn<<1];
ll n,m,N,x,y,c;
char s[3];
void build()
{
    N=1;while(N<n+2) N<<=1;
    for(int i=1;i<=n;i++)
        scanf("%lld",&sum[N+i]);
    for(int i=N-1;i;i--)
        sum[i]=sum[i<<1]+sum[i<<1|1];
}
void updata(int L,int R,int val)
{
    ll i,j,Ln=0,Rn=0,x=1;
    for(i=N+L-1,j=N+R+1;i^j^1;i>>=1,j>>=1,x<<=1)
    {
        sum[i]+=val*Ln;
        sum[j]+=val*Rn;
        if(~i&1) add[i^1]+=val,sum[i^1]+=val*x,Ln+=x;
        if(j&1) add[j^1]+=val,sum[j^1]+=val*x,Rn+=x;
    }
    for(;i;i>>=1)
    {
        sum[i]+=Ln*val;
        sum[j]+=Rn*val;
    }
}
ll query(int L,int R)
{
    ll i,j,Rn=0,Ln=0,x=1,ans=0;
    for(i=N+L-1,j=N+R+1;i^j^1;i>>=1,j>>=1,x<<=1)
    {
        if(add[i]) ans+=add[i]*Ln;
        if(add[j]) ans+=add[j]*Rn;
        if(~i&1) ans+=sum[i^1],Ln+=x;
        if(j&1) ans+=sum[j^1],Rn+=x;
    }
    for(;i;i>>=1)
    {
        ans+=add[i]*Ln;
        ans+=add[j]*Rn;
    }
    return ans;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    build();
    while(m--)
    {
        scanf("%s",&s);
        if(s[0]=='Q')
        {
            scanf("%lld%lld",&x,&y);
            printf("%lld\n",query(x,y));
        }
        else
        {
            scanf("%lld%lld%lld",&x,&y,&c);
            updata(x,y,c);
        }
    }
    return 0;
}