线段树 POJ 3468 A Simple Problem with Integers

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 abc" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q ab" 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

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;long long a[100217*4];long long col[100217*4];int temp=1;void build(int rt, int begin, int end){    int m=(begin+end)/2;    if(begin==end)    {        scanf("%I64d",&a[rt]);        return ;    }    build( rt << 1, begin, m);    build( 1 + (rt << 1), m + 1,end);    a[rt] = a[rt << 1] + a[1 + (rt << 1)];}void update(int rt, int begin, int end, int x, int y, int v){    int m=(begin+end)/2;    if(x>end||y<begin)        return ;    if(end<=y&&begin>=x)    {        a[rt]+=(end-begin+1)*v;        col[rt]+=v;        //printf("%d %d %d %d %d\n",col[rt], begin, end, rt, a[rt]);        return ;    }    if(col[rt]!=0)    {        a[rt<<1] += (m-begin+1)*col[rt];        a[(rt<<1)+1] += (end-(m+1)+1)*col[rt];        col[rt<<1] += col[rt] ;        col[(rt<<1)+1] += col[rt] ;        col[rt] = 0;    }    update( rt << 1, begin, m, x, y, v);    update( 1 + (rt << 1), m + 1,end, x, y, v);    a[rt] = a[rt << 1] + a[1 + (rt << 1)];}long long query(int rt, int begin, int end, int x, int y){    int m=(begin+end)/2;    if(x>end||y<begin)        return 0;    if(end<=y&&begin>=x)    {        //printf("%d-\n",rt);        //printf("%d\n",a[rt]);       long long t=a[rt];        return t;    }    if(col[rt]!=0)    {        a[rt<<1 ]+= (m-begin+1)*col[rt];        a[(rt<<1)+1] += (end-(m+1)+1)*col[rt];        col[rt<<1] += col[rt] ;        col[(rt<<1)+1] += col[rt] ;        col[rt] = 0;    }    long long p1 = 0,p2 = 0;    p1 = query( rt << 1, begin, m, x, y);    p2 = query( 1 + (rt << 1), m + 1,end, x, y);    //printf("%d %d\n",p1,p2);    return p1+p2;    //a[rt]=a[rt << 1] + a[1 + (rt << 1)];}int main(){    int T,n,m;    while(scanf("%d%d",&n,&m)!=-1)    {        build(1,1,n);       // puts("1");        memset(col,0,sizeof(col));       // getchar();       // printf("%d\n",a[1]);       while(m--)       {           //getchar();           int x,y,v;           char ch[10];           scanf("%s",&ch);           //printf("%c\n",ch);           if(ch[0]=='Q')           {               scanf("%d%d", &x, &y);               printf("%lld\n", query(1,1,n,x,y));              // puts("y");//               for(int i=1;i<n*4;i++)//                    printf("%I64d %d\n",a[i],i);           }           else           {               //puts("x");               scanf("%d%d%d", &x, &y, &v);                update(1,1,n,x,y,v);                //printf("%d\n",a[1]);//                for(int i=1;i<n*4;i++)//                    printf("%I64d %d\n",a[i],i);           }       }    }    return 0;}