POJ

题目描述

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.

输入

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.

输出

You need to answer all Q commands in order. One answer in a line.

样例输入

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

样例输出

4
55
9
15

提示

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

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思路

基础线段树区间更新维护区间和。
注意：
因为数据关系，所以sum数组要用long long，而且，query函数返回值也要用long long，query函数里定义的ans变量也要用long long 。
别问我怎么知道的。
先面给出我的代码，其实可以写一个结构体，不过这样写也无所谓对吧。

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代码

#include <stdio.h>#include <iostream>#include <algorithm>#include <math.h>#include <cstring>#include <queue>#include <map>#include <vector>#include <string>#define mem(a) memset(a,0,sizeof(a))#define mem2(a) memset(a,-1,sizeof(a))#define mod 1000000007#define mx 100005using namespace std;long long a[mx],sum[mx<<2],add[mx<<2];//直接开long long 就好了。sum和add数组要开数据量的四倍别忘了。int n,l;void pushup (int x)//维护当前线段的区间和。{    sum[x]=sum[x<<1]+sum[x<<1|1];}void build (int l,int r,int x)//建树{    if(l==r)    {        sum[x]=a[l];        return;    }    int mid = (l+r)/2;    build(l,mid,x<<1);    build(mid+1,r,x<<1|1);    pushup(x);}void pushdown (int x,int lx,int rx)//下推懒惰标记。{//lx,rx分别是左子树，右子树的数量。    if(add[x])    {        //下推标记        add[x<<1]+=add[x];        add[x<<1|1]+=add[x];        //修改子节点的sum使之与对应的add相对应。        sum[x<<1]+=add[x]*lx;        sum[x<<1|1]+=add[x]*rx;        //清除该节点的懒惰标记。        add[x]=0;    }}void update (int L,int R,int C,int l,int r,int x){    if(L<=l&&r<=R)    {        sum[x]+=C*(r-l+1);        add[x]+=C;        return ;    }    int mid=(l+r)/2;    pushdown(x,mid-l+1,r-mid);    //一定要在向下更新操作进行之前下推懒惰标记。    if(L<=mid) update(L,R,C,l,mid,x<<1);    if(R>mid) update(L,R,C,mid+1,r,x<<1|1);    pushup(x);}long long query (int L,int R,int l,int r,int x)//返回值一定要写long long 别问我怎么知道的。{    if(L<=l&&r<=R)    {        return sum[x];    }    int mid = (l+r)>>1;    pushdown(x,mid-l+1,r-mid);//下推懒惰标记，更新左子树和右子树维护的线段的区间和的值。    long long ans=0;    if(L<=mid) ans+=query(L,R,l,mid,x<<1);    if(R>mid )   ans+=query(L,R,mid+1,r,x<<1|1);    return ans;}int main(){#ifndef ONLINE_JUDGE    freopen("1.txt","r",stdin);#endif // ONLINE_JUDGE    ios_base::sync_with_stdio(false);    cin.tie(0);    char c;    int x,y,z;    while(cin>>n>>l)    {        mem(sum);        mem(add);        mem(a);        for(int i=1; i<=n; ++i)            cin>>a[i];        build(1,n,1);        while(l--)        {            cin>>c;            if(c=='Q')            {                cin>>x>>y;                cout<<query(x,y,1,n,1)<<endl;            }            if(c=='C')            {                cin>>x>>y>>z;                update(x,y,z,1,n,1);            }        }    }    return 0;}