[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, ... , 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
 

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 }

 

 

 

posted @ 2014-11-27 11:10  SXISZERO  阅读(193)  评论(0)  编辑  收藏  举报