POJ3468 A Simple Problem with Integers(线段树区间更新，lazy标记)

题目：

A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 114345 Accepted: 35470Case 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 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

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思路：

这一题的意思是给了一个区间，然后有两种操作，'C'可以对某段区间的所有数加上一个值，'Q'代表查询一段区间的总和，一道lazy标记的模板题

代码：

#include <cstdio> #include <cstring> #include <cctype> #include <string> #include <set> #include <iostream> #include <stack> #include <cmath> #include <queue> #include <vector> #include <algorithm> #define mem(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f #define N 100050 #define ll long long using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 ll sum[N<<2],lazy[N<<2]; void pushup(ll rt) { sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void pushdown(ll rt,ll m) { if(lazy[rt]) { lazy[rt<<1]+=lazy[rt]; lazy[rt<<1|1]+=lazy[rt]; sum[rt<<1]+=lazy[rt]*(m-(m>>1)); sum[rt<<1|1]+=lazy[rt]*(m>>1); lazy[rt]=0; } } void build(ll l,ll r,ll rt) { lazy[rt]=0; if(l==r) { scanf("%lld",&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) { lazy[rt]+=c; sum[rt]+=(ll)c*(r-l+1); return; } pushdown(rt,r-l+1); 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); ll m=(l+r)>>1; ll ans=0; if(L<=m) ans+=query(L,R,lson); if(R>m) ans+=query(L,R,rson); return ans; } int main() { ll n,m,a,b,c; char s[5]; scanf("%lld%lld",&n,&m); build(1,n,1); while(m--) { scanf("%s",s); if(s[0]=='Q') { scanf("%lld%lld",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } if(s[0]=='C') { scanf("%lld%lld%lld",&a,&b,&c); update(a,b,c,1,n,1); } } return 0; }