[POJ] 3468 A Simple Problem with Integers [线段树区间更新求和]

A Simple Problem with Integers

 

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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

 

线段树功能:update:成段增减 query:区间求和

 

  1 #include<cstdio>

  2 #include<algorithm>

  3 

  4 #define clr(x,y) memset(x,y,sizeof(x))

  5 #define LL       long long

  6 #define lson l,m,rt<<1

  7 #define rson m+1,r,rt<<1|1

  8 

  9 const int maxn=1e5+3511;

 10 using namespace std;

 11 

 12 LL sum[maxn<<2],Lazy[maxn<<2];

 13 

 14 void PushUp(int rt)

 15 {

 16     sum[rt]=sum[rt<<1]+sum[rt<<1|1];

 17 }

 18 

 19 void PushDown(int rt,int m)

 20 {

 21     if(Lazy[rt]) {

 22         Lazy[rt<<1]+=Lazy[rt];

 23         Lazy[rt<<1|1]+=Lazy[rt];

 24         sum[rt<<1]+=(m-(m>>1))*Lazy[rt];

 25         sum[rt<<1|1]+=(m>>1)*Lazy[rt];

 26         Lazy[rt]=0;

 27     }    

 28 }

 29 

 30 void build(int l,int r,int rt)

 31 {

 32     int m;

 33     Lazy[rt]=0;

 34     if(l==r) {

 35         scanf("%lld",&sum[rt]);

 36         return;

 37     }

 38     

 39     m=(l+r)>>1;

 40     build(lson);

 41     build(rson);

 42     PushUp(rt);

 43 }

 44 

 45 void Updata(int L,int R,int c,int l,int r,int rt)

 46 {

 47     int m;

 48     if(L<=l && r<=R) {

 49         Lazy[rt]+=c;

 50         sum[rt]+=(LL)c*(r-l+1);

 51         return;

 52     }

 53     

 54     PushDown(rt,r-l+1);

 55     m=(l+r)>>1;

 56     if(L<=m) Updata(L,R,c,lson);

 57     if(R>m)  Updata(L,R,c,rson);

 58     PushUp(rt);

 59     

 60 }

 61 

 62 

 63 LL query(int L,int R,int l,int r,int rt)

 64 {

 65     int m;

 66     LL ret=0;

 67     if(L<=l && r<=R) {

 68         return sum[rt];

 69     }

 70     

 71     PushDown(rt,r-l+1);

 72     m=(l+r)>>1;

 73     if(L<=m) ret+=query(L,R,lson);

 74     if(R>m)  ret+=query(L,R,rson);

 75     

 76     return  ret;

 77 }

 78 

 79 

 80 

 81 int main()

 82 {

 83     int Q,n,a,b,c;

 84     char st[10];

 85     

 86     scanf("%d%d",&n,&Q);

 87     build(1,n,1);

 88     

 89     while(Q--) {

 90         scanf("%s",st);

 91         if(st[0]=='C') {

 92             scanf("%d%d%d",&a,&b,&c);

 93             Updata(a,b,c,1,n,1);

 94         } else {

 95             scanf("%d%d",&a,&b);

 96             printf("%lld\n",query(a,b,1,n,1));

 97         }

 98         

 99     }

100     

101     return 0;

102 }