poj-3468A Simple Problem with Integers（线段树对部分数值的改变以及求和）

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 96612 Accepted: 30145Case 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

#include<iostream>#include<string>#include<cstring>#include<cstdio>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define N 111111#define LL __int64using namespace std;LL add[N<<2];//保存改变值的大小 LL sum[N<<2];//保存和 void pushup(LL rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void pushdown(LL rt,LL m){if(add[rt]){add[rt<<1]+=add[rt];//左子树和右子数需要改变的总的变化量  不直接把每次的变化量传递给每个节点 只是先标记 看最后求和的时候最终改变的大小 加快速度 add[rt<<1|1]+=add[rt];sum[rt<<1]+=add[rt]*(m-(m>>1));//改变左子树节点的sum值 sum[rt<<1|1]+=add[rt]*(m>>1);add[rt]=0;//清楚该节点的标记 }}void build(LL l,LL r,LL rt){add[rt]=0;if(l==r){scanf("%I64d",&sum[rt]);return;}LL m=(l+r)>>1;build(lson);build(rson);pushup(rt);}void update(LL L,LL R,LL c,LL l,LL r,LL rt){if(L<=l&&R>=r){add[rt]+=c;sum[rt]+=(LL)c*(r-l+1);//r-l+1表示r到l区间的节点数量 return ;}pushdown(rt,r-l+1);//更新子节点 LL m=(l+r)>>1;if(L<=m)update(L,R,c,lson);if(m<R)update(L,R,c,rson);pushup(rt);}LL query(LL L,LL R,LL l,LL r,LL rt){if(L<=l&&R>=r){return sum[rt];}pushdown(rt,r-l+1);// 把标记下推 跟新到每个sum 不然会出错 LL m=(l+r)>>1;LL res=0;if(m>=L){res+=query(L,R,lson);}if(m<R){res+=query(L,R,rson);}return res;}int main(){LL n,q;scanf("%I64d%I64d",&n,&q);build(1,n,1);while(q--){char op[2];LL a,b,c;scanf("%s",op);if(op[0]=='Q'){scanf("%I64d%I64d",&a,&b);printf("%I64d\n",query(a,b,1,n,1));} else {scanf("%I64d%I64d%I64d",&a,&b,&c);update(a,b,c,1,n,1);}}return 0;}

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