POJ 3468 A Simple Problem with Integers （线段树 区间共加）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 112230 Accepted: 34906Case 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

线段树的区间更改，求区间和。

注意用ll，数据超int

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>using namespace std;const int N = 100440*8;#define LL long longLL sum[N],add[N];LL a[N/5];void build(int x,int l,int r){    if(l==r) sum[x]=a[l];    else    {        int mid=(l+r)>>1;        build(2*x,l,mid);        build(2*x+1,mid+1,r);        sum[x]=sum[x*2]+sum[x*2+1];    }}void pushdown(int x,int l,int r){    int mid=(l+r)>>1;    add[2*x]+=add[x];    sum[2*x]+=(LL)(mid-l+1)*add[x];    add[2*x+1]+=add[x];    sum[2*x+1]+=(LL)(r-mid)*add[x];    add[x]=0;}void Add(int x,int l,int r,int ll,int rr,LL c){    if(add[x]!=0) pushdown(x,l,r);    if(ll<=l&&rr>=r)    {        sum[x]+=(LL)(r-l+1)*c;        add[x]+=(LL)c;    }    else    {        int mid=(l+r)>>1;        if(ll<=mid) Add(2*x,l,mid,ll,rr,c);        if(mid<rr) Add(2*x+1,mid+1,r,ll,rr,c);        sum[x]=sum[x*2]+sum[x*2+1];    }    }LL query(int x,int l,int r,int ll,int rr){    if(add[x]!=0) pushdown(x,l,r);    LL ans=0;    if(ll<=l&&rr>=r) ans+=sum[x];    else    {        int mid=(l+r)>>1;        if(ll<=mid) ans+=query(2*x,l,mid,ll,rr);        if(mid<rr) ans+=query(2*x+1,mid+1,r,ll,rr);    }    return ans;}int main(){    int n,q;    while(~scanf("%d %d",&n,&q))    {        memset(add,0,sizeof add);        for(int i=1;i<=n;i++)        {            scanf("%lld",&a[i]);        }        build(1,1,n);        while(q--)        {            char s;            getchar();            scanf("%c",&s);            if(s=='C')            {                int l,r;                LL c;                scanf("%d %d %lld",&l,&r,&c);                Add(1,1,n,l,r,c);//                for(int i=1;i<=n;i++)//                {//                    printf("%lld ",query(1,1,n,i,i));//                }//                printf("\n");            }            else            {                int l,r;                scanf("%d %d",&l,&r);                LL ans=query(1,1,n,l,r);                printf("%lld\n",ans);            }        }    }}