POJ 3468 线段树 解题报告

A Simple Problem with Integers

Description

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 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

Sample Output

4
55
9
15

【解题报告】
区间修改区间查询线段树。
写一下练练手

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 100010#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1long long add[maxn<<2],sum[maxn<<2];int n,q;void pushup(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];    }  void pushdown(int rt,int m){    if(add[rt])    {        add[rt<<1]+=add[rt];        add[rt<<1|1]+=add[rt];        sum[rt<<1]+=add[rt]*(m-(m>>1));        sum[rt<<1|1]+=add[rt]*(m>>1);        add[rt]=0;    }}void build(int l,int r,int rt){    add[rt]=0;    if(l==r)    {        scanf("%lld",&sum[rt]);        return;    }    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int L,int R,int c,int l,int r,int rt){    if(L<=l&&R>=r)    {        add[rt]+=(long long)c;        sum[rt]+=(long long)c*(r-l+1);        return;    }    pushdown(rt,r-l+1);    int m=(r+l)>>1;    if(L<=m) update(L,R,c,lson);    if(R>m) update(L,R,c,rson);    pushup(rt);}long long query(int L,int R,int l,int r,int rt){    if(L<=l&&R>=r)    {        return sum[rt];    }    pushdown(rt,r-l+1);     int m=(l+r)>>1;    long long ret=0;    if(L<=m) ret+=query(L,R,lson);    if(R>m) ret+=query(L,R,rson);    return ret;}int main(){    scanf("%d%d",&n,&q);    build(1,n,1);    while(q--)    {        char opt[5];        int a,b,c;        scanf("%s",opt);        if(opt[0]=='Q')         {            scanf("%d%d",&a,&b);            printf("%lld\n",query(a,b,1,n,1));        }        if(opt[0]=='C')        {            scanf("%d%d%d",&a,&b,&c);            update(a,b,c,1,n,1);        }    }    return 0;}