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 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

Hint

The sums may exceed the range of 32-bit integers.

线段树的区间更新和区间和的查询，加了lazy的优化，用的还是不熟练，太菜。。。有几处错误gg，这里标注一下，以后长点记性：

       ( 1 ) 跟set不一样，这里用的是add，是在给定区间上的每个数加上一个数，pushdown里s[lc]和s[rc]加的是下推了的lazy(s[d].lazy)，而不是lc和rc的lazy(s[lc).lazy、s[rc].lazy)，（s[lc].sum += (s[d].lazy * (m - l + 1));  s[rc].sum += (s[d].lazy * (r - m));）。

       ( 2 ) query里不需要pushup；

代码如下：

#include<cstdio>#include<algorithm>#include<cstring>#define ll long longusing namespace std;typedef struct node{    ll sum,lazy;}node;const int maxn = 100000 + 5;const int INF = 1 << 30;node s[maxn << 4];ll a[maxn];void pushdown(int d,int l,int r){    int lc = d * 2,rc = d * 2 + 1,m = l + (r - l) / 2;    s[lc].lazy += s[d].lazy;    s[rc].lazy += s[d].lazy;    s[lc].sum += (s[d].lazy * (m - l + 1));    s[rc].sum += (s[d].lazy * (r - m));    s[d].lazy = 0;}void pushup(int d,int l,int r){    s[d].sum = s[d * 2].sum + s[d * 2 + 1].sum;}void build(int d,int l,int r){    s[d].lazy = 0;    if(l == r)    {        s[d].sum = a[l];        return;    }    int m = l + (r - l) / 2;    build(d * 2,l,m);    build(d * 2 + 1,m + 1,r);    pushup(d,l,r);}void update(int d,int x,int y,int l,int r,int v){    if(x <= l && r <= y)    {        s[d].lazy += v;        s[d].sum += (v*(r - l + 1));        return;    }    int m = l +(r - l) / 2;    if(s[d].lazy) pushdown(d,l,r);    if(x <= m) update(d * 2,x,y,l,m,v);    if(m < y) update(d * 2 + 1,x,y,m + 1,r,v);    pushup(d,l,r);}ll query(int d,int x,int y,int l,int r){    ll ss = 0;    if(x <= l && r <= y)    {        ss += s[d].sum;        return ss;    }    int m = l + (r - l) / 2;    if(s[d].lazy) pushdown(d,l,r);    if(x <= m) ss += query(d * 2,x,y,l,m);    if(m < y) ss += query(d * 2 + 1,x,y,m + 1,r);    return ss;}int main(){    int n,q,x,y,v;    char p;    scanf("%d %d",&n,&q);    for(int i = 1;i <= n; ++i) scanf("%lld",&a[i]);    build(1,1,n);    for(int i = 0;i < q; ++i)    {        scanf(" %c",&p);        if(p == 'Q')        {            scanf("%d %d",&x,&y);            printf("%lld\n",query(1,x,y,1,n));        }        else if(p == 'C')        {            scanf("%d %d %d",&x,&y,&v);            update(1,x,y,1,n,v);        }    }    return 0;}