POJ-3468A Simple Problem with Integers(线段树区间修改)

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 116429 Accepted: 36174
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.
Source

POJ Monthly–2007.11.25, Yang Yi

题意：

需要支持两种操作，区间求和，区间修改
线段树加lazy标记，就当写个版

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn=2e5+5;int a[maxn]; LL T[4*maxn],tag[4*maxn],z; char opt[5];int L,R,x,y,n,m;inline void down(int l,int r,int k){  tag[k<<1]+=tag[k];tag[k<<1|1]+=tag[k];  int mid=(l+r)>>1;  T[k<<1]+=(mid-l+1)*tag[k];  T[k<<1|1]+=(r-mid)*tag[k];  tag[k]=0;}inline void update(int k){  T[k]=T[k<<1]+T[k<<1|1];}inline void build(int l,int r,int k){  if(l==r)  {    T[k]=a[l];    tag[k]=0;    return;  }  int mid=(l+r)>>1;  build(l,mid,k<<1);  build(mid+1,r,k<<1|1);  update(k);  tag[k]=0;}inline LL ask(int l,int r,int k){  if(L<=l&&r<=R)return T[k];  int mid=(l+r)>>1;  down(l,r,k);  LL res=0;  if(L<=mid)res+=ask(l,mid,k<<1);  if(R>mid)res+=ask(mid+1,r,k<<1|1);  return res;  }inline void add(int l,int r,int k){  if(L<=l&&r<=R)  {    tag[k]+=z;    T[k]+=(r-l+1)*z;    return ;  }  int mid=(l+r)>>1;  down(l,r,k);  if(L<=mid)add(l,mid,k<<1);  if(R>mid)add(mid+1,r,k<<1|1);  update(k);}int main(){  while(~scanf("%d%d",&n,&m))  {    for(int i=1;i<=n;++i)scanf("%d",a+i);    build(1,n,1);    for(int i=1;i<=m;++i)    {      scanf("%s%d%d",opt,&x,&y);      if(opt[0]=='Q')      {        L=x,R=y;        printf("%lld\n",ask(1,n,1));      }      else      {        scanf("%lld",&z);        L=x,R=y;        add(1,n,1);      }    }  }}