POJ 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 48592 Accepted: 14381Case 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

  线段树区间更新

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#define N 100010using namespace std;struct num{    int l,r;    __int64 val,add,sum;}a[4*N];__int64 b[N],res;bool status[10*N];char s1[5];int main(){    //freopen("data.in","r",stdin);    void pre_build(int k,int l,int r);    void find(int k,int l,int r);    void update(int k,int l,int r,__int64 val);    int n,m;    while(scanf("%d %d",&n,&m)!=EOF)    {        memset(status,false,sizeof(status));        for(int i=1;i<=n;i++)        {            scanf("%I64d",&b[i]);        }        pre_build(1,1,n);        while(m--)        {           scanf("%s",s1);           int x,y;           __int64 val;           if(s1[0]=='Q')           {               scanf("%d %d",&x,&y);               res=0;               find(1,x,y);               printf("%I64d\n",res);           }else           {               scanf("%d %d %I64d",&x,&y,&val);               update(1,x,y,val);           }        }    }    return 0;}void pushup(int k){    a[k].sum=a[k<<1].sum+a[k<<1|1].sum;}void pre_build(int k,int l,int r){    a[k].l = l;    a[k].r = r;    a[k].val=0;    a[k].add=0;    status[k]=true;    if(l==r)    {        a[k].sum=b[l];        return ;    }    int mid=(l+r)>>1;    pre_build(k<<1,l,mid);    pre_build(k<<1|1,mid+1,r);    pushup(k);}void update(int k,int l,int r,__int64 val){    a[k].sum+=(__int64)(r-l+1)*val;    if(a[k].l==l&&a[k].r==r)    {        a[k].val+=val;        return ;    }    int mid=(a[k].l+a[k].r)>>1;    if(r<=mid)    {        update(k<<1,l,r,val);    }else if(l>mid)    {        update(k<<1|1,l,r,val);    }else    {        update(k<<1,l,mid,val);        update(k<<1|1,mid+1,r,val);    }}void find(int k,int l,int r){    a[k].sum+=(a[k].add*(__int64)(a[k].r-a[k].l+1));    if(status[k<<1])    {        a[k<<1].add+=(a[k].add+a[k].val);    }    if(status[k<<1|1])    {        a[k<<1|1].add+=(a[k].add+a[k].val);    }    a[k].add=0;    a[k].val=0;    if(a[k].l==l&&a[k].r==r)    {        res+=a[k].sum;        return ;    }    int mid=(a[k].l+a[k].r)>>1;    if(r<=mid)    {        find(k<<1,l,r);    }else if(l>mid)    {        find(k<<1|1,l,r);    }else    {        find(k<<1,l,mid);        find(k<<1|1,mid+1,r);    }}