poj 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K   Case 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

Hint

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

解题思路：线段树区间更新模板题。

链接：http://poj.org/problem?id=3468

代码：

#include <iostream>#define N 100005#include <stdio.h>using namespace std;long long num[N];struct Tree{    int l,r;long long sum;    long long add;}tree[N*4];void PushUp(int root){    tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;}void PushDown(int root,int m){    if(tree[root].add)    {        tree[root<<1].add+=tree[root].add;        tree[root<<1|1].add+=tree[root].add;        tree[root<<1].sum+=(m-(m>>1))*tree[root].add;        tree[root<<1|1].sum+=(m>>1)*tree[root].add;        tree[root].add=0;    }}void build(int root,int l,int r){    tree[root].l=l;    tree[root].r=r;    tree[root].add=0;    if(tree[root].l==tree[root].r)    {        tree[root].sum=num[l];        return;    }    int mid=(l+r)/2;    build(root<<1,l,mid);    build(root<<1|1,mid+1,r);    PushUp(root);}void update(int L,int R,int c,int l,int r,int root){    if(l>=L&&r<=R)    {        tree[root].add+=c;        tree[root].sum+=(long long)c*(r-l+1);        return ;    }    PushDown(root,r-l+1);    int mid=(l+r)/2;    if(mid>=L)    update(L,R,c,l,mid,root<<1);    if(mid<R)    update(L,R,c,mid+1,r,root<<1|1);    PushUp(root);}long long query(int L,int R,int l,int r,int root){    if(L<=l&&R>=r)    return tree[root].sum;    PushDown(root,r-l+1);    long long ret=0;    int mid=(l+r)>>1;    if(mid>=L) ret+=query(L,R,l,mid,root<<1);    if(mid<R) ret+=query(L,R,mid+1,r,root<<1|1);    return ret;}int main(){    int m,n;    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    scanf("%I64d",&num[i]);    build(1,1,n);    while(m--)    {        char s[6];        int a,b,c;        scanf("%s",s);        if(s[0]=='Q')        {            scanf("%d%d",&a,&b);            printf("%I64d\n",query(a,b,1,n,1));        }        else        {            scanf("%d%d%d",&a,&b,&c);            update(a,b,c,1,n,1);        }    }    return 0;}