A Simple Problem with Integers POJ

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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint
The sums may exceed the range of 32-bit integers.
题意：对这样一组数进行下列操作： Q操作表示读取该区间上的和，C操作表示为i到j区间的数全部增加c。

#include<stdio.h>#include<string.h>#include<stdalign.h>int num[400000];struct node{    int l,r;    long long nsum;    long long add;}seg[400000];void Build(int i,int l,int r)//建立线段树{    seg[i].l=l;    seg[i].r=r;    seg[i].add=0;    if(l==r)    {        seg[i].nsum = num[l];        return ;    }    int mid = (l+r)/2;    Build(i*2,l,mid);    Build(i*2+1,mid+1,r);    seg[i].nsum = seg[i*2].nsum+seg[i*2+1].nsum;}void Add(int i,int l,int r,long long c)//区间上的每个数增加{    if(seg[i].l==l&&seg[i].r==r)    {        seg[i].add+=c;        return ;    }    seg[i].nsum += c*(r-l+1);//这个区间总共增加了这些    int mid = (seg[i].l+seg[i].r)/2;    if(r<=mid)    Add(i*2,l,r,c);    else if(l>mid)    Add(i*2+1,l,r,c);    else    {        Add(i*2,l,mid,c);        Add(i*2+1,mid+1,r,c);    }}long long Query(int i,int a,int b)//查询a-b的总和{    if(seg[i].l==a&&seg[i].r==b)    {        return seg[i].nsum+(b-a+1)*seg[i].add;    }    seg[i].nsum+=(seg[i].r-seg[i].l+1)*seg[i].add;    int mid=(seg[i].l+seg[i].r)/2;    Add(i*2,seg[i].l,mid,seg[i].add);    Add(i*2+1,mid+1,seg[i].r,seg[i].add);    seg[i].add=0;    if(b<=mid)    return Query(i*2,a,b);    else if(a>mid)    return Query(i*2+1,a,b);    else    return Query(i*2,a,mid)+Query(i*2+1,mid+1,b);}int main(){    int n,m,i;    int a,b,c;    char s[10];    while(~scanf("%d%d",&n,&m))    {        for(i=1;i<=n;i++)        {            scanf("%d",&num[i]);        }        Build(1,1,n);        for(i=0;i<m;i++)        {            scanf("%s",s);            if(s[0]=='C')            {                scanf("%d%d%d",&a,&b,&c);                Add(1,a,b,c);            }            else            {                scanf("%d%d",&a,&b);                printf("%lld\n",Query(1,a,b));            }        }    }    return 0;}