bzoj 3212: Pku3468 A Simple Problem with Integers （线段树）

3212: Pku3468 A Simple Problem with Integers

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1565  Solved: 676
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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

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题解： 线段树裸题。。。。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define N 100003#define LL long long using namespace std;int n,m;LL a[N],delta[N*4],tr[N*4];void update(int now){tr[now]=tr[now<<1|1]+tr[now<<1];}void build(int now,int l,int r){if (l==r) { tr[now]=a[l]; return; }int mid=(l+r)/2;build(now<<1,l,mid);build(now<<1|1,mid+1,r);update(now);}void pushdown(int now,int l,int r){int mid=(l+r)/2;if (delta[now]){tr[now<<1]+=delta[now]*(LL)(mid-l+1);tr[now<<1|1]+=delta[now]*(LL)(r-mid);    delta[now<<1]+=delta[now];    delta[now<<1|1]+=delta[now];    delta[now]=0;}}void qjchange(int now,int l,int r,int ll,int rr,LL v){if (ll<=l&&r<=rr){delta[now]+=v; tr[now]+=v*(LL)(r-l+1);return;}int mid=(l+r)/2;pushdown(now,l,r);if (ll<=mid)  qjchange(now<<1,l,mid,ll,rr,v);if (rr>mid) qjchange(now<<1|1,mid+1,r,ll,rr,v);update(now);}LL qjsum(int now,int l,int r,int ll,int rr){if (ll<=l&&r<=rr) return tr[now];int mid=(l+r)/2;pushdown(now,l,r);LL ans=0;if (ll<=mid) ans+=qjsum(now<<1,l,mid,ll,rr);if (rr>mid) ans+=qjsum(now<<1|1,mid+1,r,ll,rr);return ans;}int main(){scanf("%d%d",&n,&m);for (int i=1;i<=n;i++) scanf("%lld",&a[i]);build(1,1,n);for (int i=1;i<=m;i++)    {      char s[10]; LL z;  int x,y;  scanf("%s%d%d",s,&x,&y);  if (s[0]=='C'){  scanf("%lld",&z);    qjchange(1,1,n,x,y,z);  }  else printf("%lld\n",qjsum(1,1,n,x,y));}}