lazy标记
  Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u 
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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 abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" 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.

<span style="color:#6633ff;">/********************************************************author    :    Grant Yuantime      :    2014.7.28algorithm :    线段树+lazy标记source    :    POJ 3468*********************************************************/#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define MAX 100003
#define LL long longusing namespace std;
struct node{
LL l;
LL r;
LL sum;
LL lazy;
};
node tree[3*MAX];
int n,m;
LL input[MAX],ans;void build(LL left,LL right,LL root)
{tree[root].lazy=0;tree[root].l=left;tree[root].r=right;if(left==right){tree[root].sum=input[left];return;}int mid=(left+right)>>1;build(left,mid,root*2);build(mid+1,right,root*2+1);tree[root].sum=tree[root*2].sum+tree[root*2+1].sum;
}void upchild(LL root)
{if(tree[root].lazy){tree[root<<1].sum+=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].lazy;tree[(root<<1)|1].sum+=(tree[(root<<1)|1].r-tree[(root<<1)|1].l+1)*tree[root].lazy;tree[root<<1].lazy+=tree[root].lazy;tree[root<<1|1].lazy+=tree[root].lazy;tree[root].lazy=0;}
}void update(LL left,LL right,LL root,LL p)
{if(left<=tree[root].l&&right>=tree[root].r){tree[root].sum+=(LL)(tree[root].r-tree[root].l+1)*p;tree[root].lazy+=(LL)p;return;}upchild(root);int mid=(tree[root].l+tree[root].r)>>1;if(mid<right)update(left,right,root*2+1,p);if(mid>=left)update(left,right,root*2,p);tree[root].sum=tree[root<<1].sum+tree[(root<<1)|1].sum;
}void query(LL left,LL right,LL root)
{if(left<=tree[root].l&&right>=tree[root].r){   ans+=(LL)tree[root].sum;return;}upchild(root);int mid=(tree[root].l+tree[root].r)>>1;if(mid>=left)query(left,right,root<<1);if(mid<right)query(left,right,root*2+1);
}int main()
{   char s;LL a,b,q;while(~scanf("%d%d",&n,&m)){for(int i=1;i<=n;i++)scanf("%lld",&input[i]);build(1,n,1);for(int i=0;i<m;i++){  getchar();scanf("%c",&s);if(s=='Q'){ans=0;scanf("%lld%lld",&a,&b);query(a,b,1);printf("%lld\n",ans);}else if(s=='C'){scanf("%lld%lld%lld",&a,&b,&q);update(a,b,1,q);}}}return 0;
}
</span>