poj 3468 A Simple Problem with Integers

poj 3468

                                                                                                 A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 47652 Accepted: 14008Case 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

代码：

#include<stdio.h>#include<string.h>#define max 100010
int w[max];
struct node{    int l,r;//区间的左右端点；    __int64 value ,add; // value节点的值（以该节点为根节点的子数的和）,add增加量   }tree[3*max];     void build(int p,int left,int right){//创建线段树； int v,mid;    tree[p].l=left; tree[p].r=right; tree[p].add=0;  if(left==right) {  tree[p].value=w[left];  return; } mid=(left+right)>>1; v=p<<1; build(v,left,mid); build(v+1,mid+1,right); tree[p].value=tree[v].value+tree[v+1].value;}
void updata(int p,int left,int right,int add){//更新区间； int v,mid; if(tree[p].l==left && tree[p].r==right) {  tree[p].add+=add;  return ; } tree[p].value+=(right-left+1)*add; v=p<<1; mid=(tree[p].l+tree[p].r)>>1; if(right<=mid)  updata(v,left,right,add); else if(left>=mid+1)  updata(v+1,left,right,add); else {  updata(v,left,mid,add);  updata(v+1,mid+1,right,add); }}
__int64 query(int p,int left,int right){//查询 int v,mid; v=p<<1; mid=(tree[p].l+tree[p].r)>>1; if(tree[p].l==left && tree[p].r==right) {  return (tree[p].value+(tree[p].r-tree[p].l+1)*tree[p].add); // return tree[p].value; } else {  tree[v].add+=tree[p].add;  tree[v+1].add+=tree[p].add;  tree[p].value+=(tree[p].r-tree[p].l+1)*tree[p].add;  tree[p].add=0; } if(right<=mid)  return query(v,left,right); else if(left>=mid+1)  return query(v+1,left,right); else  return query(v,left,mid)+query(v+1,mid+1,right);}int main(){    int N,Q,a,b,c,i;    char s[3];    while(~scanf("%d %d",&N,&Q))    {  memset(w,0,sizeof(w));        for(i=1;i<=N;i++)             scanf("%d",&w[i]);        build(1,1,N);  for(i=0;i<Q;i++)  {   scanf("%s",&s);   if(s[0]=='C')   {    scanf("%d %d %d",&a,&b,&c);    updata(1,a,b,c);   }   else   {    scanf("%d %d",&a,&b);    printf("%I64d\n",query(1,a,b));   }  } } return 0;}