A Simple Problem with Integers(线段树入门题)

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.
 
 
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <string>
  7 #include <vector>
  8 #include <stack>
  9 #include <queue>
 10 #include <set>
 11 #include <map>
 12 #include <list>
 13 #include <iomanip>
 14 #include <cstdlib>
 15 #include <sstream>
 16 using namespace std;
 17 typedef long long LL;
 18 const int INF=0x5fffffff;
 19 const double EXP=1e-6;
 20 const int MS=100005;
 21 
 22 struct node
 23 {
 24       int l,r;
 25       LL sum,inc;   //特别注意这里
 26       int mid()
 27       {
 28             return (l+r)/2;
 29       }
 30 }nodes[4*MS];
 31 
 32 void creat(int root,int l,int r)
 33 {
 34       nodes[root].l=l;
 35       nodes[root].r=r;
 36       nodes[root].sum=0;
 37       nodes[root].inc=0;
 38       if(l==r)
 39             return ;
 40       creat(root<<1,l,(l+r)/2);
 41       creat(root<<1|1,(l+r)/2+1,r);
 42 }
 43 
 44 void insert(int root,int pos,int value)
 45 {
 46       if(nodes[root].l==nodes[root].r)
 47       {
 48             nodes[root].sum+=value;
 49             return ;
 50       }
 51       nodes[root].sum+=value;
 52       if(pos<=nodes[root].mid())
 53             insert(root<<1,pos,value);
 54       else
 55             insert(root<<1|1,pos,value);
 56 }
 57 
 58 void add(int root,int a,int b,int c)
 59 {
 60       if(nodes[root].l==a&&nodes[root].r==b)
 61       {
 62             nodes[root].inc+=c;
 63             return ;
 64       }
 65       nodes[root].sum+=c*(b-a+1);
 66       if(b<=nodes[root].mid())
 67             add(root<<1,a,b,c);
 68       else if(a>nodes[root].mid())
 69             add(root<<1|1,a,b,c);
 70       else
 71       {
 72             add(root<<1,a,nodes[root].mid(),c);
 73             add(root<<1|1,nodes[root].mid()+1,b,c);
 74       }
 75 }
 76 
 77 LL  query(int root,int a,int b)
 78 {
 79       if(nodes[root].l>=a&&nodes[root].r<=b)
 80             return nodes[root].sum+nodes[root].inc*(nodes[root].r-nodes[root].l+1);
 81       nodes[root].sum+=nodes[root].inc*(nodes[root].r-nodes[root].l+1);
 82       add(root<<1,nodes[root].l,nodes[root].mid(),nodes[root].inc);
 83       add(root<<1|1,nodes[root].mid()+1,nodes[root].r,nodes[root].inc);
 84       nodes[root].inc=0;
 85       if(b<=nodes[root].mid())
 86             return query(root<<1,a,b);
 87       else if(a>nodes[root].mid())
 88             return query(root<<1|1,a,b);
 89       else
 90             return query(root<<1,a,nodes[root].mid())+query(root<<1|1,nodes[root].mid()+1,b);
 91 }
 92 
 93 int main()
 94 {
 95       int N,Q,x,y,z;
 96       scanf("%d%d",&N,&Q);
 97       creat(1,1,N);
 98       for(int i=1;i<=N;i++)
 99       {
100             scanf("%d",&x);
101             insert(1,i,x);
102       }
103       char cmd[MS];
104       while(Q--)
105       {
106             scanf("%s",cmd);
107             if(cmd[0]=='Q')
108             {
109                   scanf("%d%d",&x,&y);
110                   printf("%lld\n",query(1,x,y));
111             }
112             else
113             {
114                   scanf("%d%d%d",&x,&y,&z);
115                   add(1,x,y,z);
116             }
117       }
118       return 0;
119 }

 

 
 

 

posted @ 2014-07-22 22:40  daydaycode  阅读(1015)  评论(0)  编辑  收藏  举报