poj 3468 A Simple Problem with Integers

http://poj.org/problem?id=3468

A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 38088   Accepted: 11046
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


成段更新，区间求和。。。入门题。。
为嘛我做了这么久，还要照着模版打。。。。弱。
View Code 
  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #define maxn 100010
  4 #define lll   __int64
  5 lll num[maxn];
  6 lll ans;
  7 struct Node
  8 {
  9    lll sum,p;
 10    int  l,r;
 11 }node[maxn*4];
 12 void build_tree(int rt,int ll,int rr)
 13 {
 14       node[rt].l=ll;
 15       node[rt].r=rr;
 16       node[rt].p=0;
 17       if(ll==rr)
 18       {
 19             node[rt].sum=num[ll];
 20             
 21             return;
 22       }
 23       int mid=(node[rt].l+node[rt].r)/2;
 24       build_tree(rt*2,ll,mid);
 25       build_tree(rt*2+1,mid+1,rr);
 26       node[rt].sum=node[rt*2].sum+node[rt*2+1].sum;
 27 }
 28 
 29 void update(int rt,int s,int e,lll value)
 30 {
 31     if(node[rt].l==s&&node[rt].r==e)
 32     {
 33           node[rt].sum+=(e-s+1)*value;
 34           node[rt].p+=value;
 35           return;
 36     }
 37     if(node[rt].p)
 38     {
 39         node[rt*2].p+=node[rt].p;
 40             node[rt*2+1].p+=node[rt].p;
 41             node[rt*2].sum+=(node[rt*2].r-node[rt*2].l+1)*node[rt].p;
 42             node[rt*2+1].sum+=(node[rt*2+1].r-node[rt*2+1].l+1)*node[rt].p;
 43             node[rt].p=0;
 44     }
 45     int mid=(node[rt].l+node[rt].r)/2;
 46     if(e<=mid)
 47     {
 48           update(rt*2,s,e,value);
 49     }
 50     else if(s>mid)
 51     {
 52           update(rt*2+1,s,e,value);
 53     }
 54     else
 55     {
 56           update(rt*2,s,mid,value);
 57           update(rt*2+1,mid+1,e,value);
 58     }
 59     node[rt].sum=node[rt*2].sum+node[rt*2+1].sum;
 60 }
 61 
 62 void query(int rt,int x,int y)
 63 {
 64       if(node[rt].l==x&&node[rt].r==y)
 65       {
 66             ans+=node[rt].sum;
 67             return;
 68       }
 69       else
 70       {
 71             node[rt*2].p+=node[rt].p;
 72             node[rt*2+1].p+=node[rt].p;
 73             node[rt*2].sum+=(node[rt*2].r-node[rt*2].l+1)*node[rt].p;
 74             node[rt*2+1].sum+=(node[rt*2+1].r-node[rt*2+1].l+1)*node[rt].p;
 75             node[rt].p=0;
 76       }
 77       int mid=(node[rt].l+node[rt].r)/2;
 78       if(y<=mid)
 79       {
 80             query(rt*2,x,y);
 81       }
 82       else if(x>mid)
 83       {
 84             query(rt*2+1,x,y);
 85       }
 86       else
 87       {
 88             query(rt*2,x,mid);
 89             query(rt*2+1,mid+1,y);
 90       }
 91 
 92 }
 93 int main()
 94 {
 95     lll val;
 96     int n,q;
 97     int i,x,y;
 98     char ch;
 99     while(~scanf("%d%d",&n,&q))
100     {
101           for(i=1;i<=n;i++)
102           {
103                 scanf("%I64d",&num[i]);
104           }
105           build_tree(1,1,n);
106           for(i=0;i<q;i++)
107           {
108                 getchar();
109                 scanf("%c",&ch);
110                 if(ch=='C')
111                 {
112                       scanf("%d%d%I64d",&x,&y,&val);
113                       update(1,x,y,val);
114                 }
115                 else if(ch=='Q')
116                 {
117                       ans=0;
118                       scanf("%d%d",&x,&y);
119                       query(1,x,y);
120                       printf("%I64d\n",ans);
121                 }
122           }
123     }
124 }

 

posted on 2012-11-15 22:19  仁者无敌8勇者无惧  阅读(197)  评论(0)  编辑  收藏  举报

导航