A Simple Problem with Integers

Time Limit: 5000MSMemory Limit: 131072KTotal Submissions: 107094Accepted: 33429Case Time Limit: 2000MS

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 线段树模板题（虽然我还是不会写 首先数据量比较大 要用LL 直接套模板 修改单个点会T不动脑子 看了看dalao的博客 改了改代码 解析在代码内2333 #include <cstdio> #include <algorithm> using namespace std; #define LL long long const int N = 1000000*4+10; int arr[N]; struct node { int left, right; LL sum, num; }tree[N]; void build(int l,int r,int step) { tree[step].left = l; tree[step].right = r; tree[step].num = 0; if(l == r) { tree[step].sum = arr[l]; return ; } int mid = (l+r) >> 1; build(l,mid,step<<1); build(mid+1,r,step<<1|1); tree[step].sum = tree[step<<1].sum + tree[step<<1|1].sum; } void update(int l,int r,LL value,int step) { if(tree[step].left==l&&r==tree[step].right) { tree[step].num += value; return; } tree[step].sum += value*(r-l+1); int mid = (tree[step].left+tree[step].right) >> 1; if(r <= mid) update(l,r,value,step<<1); else if(l > mid) update(l,r,value,step<<1|1); else { update(l,mid,value,step<<1); update(mid+1,r,value,step<<1|1); } } LL query(int l,int r,int step) { if(l==tree[step].left && r==tree[step].right) { return tree[step].sum+(r-l+1)*tree[step].num; } int mid = (tree[step].left+tree[step].right) >> 1; tree[step].sum += (tree[step].right-tree[step].left+1)*tree[step].num; update(tree[step].left,mid,tree[step].num,step<<1); update(mid+1,tree[step].right,tree[step].num,step<<1|1); tree[step].num = 0; if(r <= mid) return query(l,r,step<<1); else if(l > mid) return query(l,r,step<<1|1); else return query(l,mid,step<<1)+query(mid+1,r,step<<1|1); } int main() { int n, q, x; scanf("%d%d",&n,&q); for(int i = 1;i <= n; i++) { scanf("%d",&arr[i]); } build(1,n,1); while(q--) { getchar(); char ch; int u, v, z; scanf("%c",&ch); if(ch == 'C') { scanf("%d%d%d",&u,&v,&z); update(u,v,z,1); } if(ch == 'Q') { scanf("%d%d",&u,&v); printf("%lld\n",query(u,v,1)); } } return 0; }