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
Hint
The sums may exceed the range of 32-bit integers.
题意：区间修改与查询区间和
分析
线段树的基本操作，使用lazy标记的思想。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<algorithm> #include<cstring> #include <queue> #include <vector> #include<bitset> #include<map> #include<deque> using namespace std; typedef long long LL; const int maxn = 1e5+150; const int mod = 77200211+233; typedef pair<int,int> pii; #define X first #define Y second #define pb push_back #define mp make_pair #define ms(a,b) memset(a,b,sizeof(a)) const int inf = 0x3f3f3f3f; #define lson l,m,2*rt #define rson m+1,r,2*rt+1 LL sum[maxn<<2],lazy[maxn<<2]; struct node{ int l,r; int mid(){ return (l+r)>>1; } }tree[maxn<<2]; //更新父亲信息 void PushUp(int rt){ sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } //更新子孙信息 void PushDown(int rt,int len){ if(lazy[rt]){ lazy[rt<<1]+=lazy[rt]; lazy[rt<<1|1]+=lazy[rt]; sum[rt<<1] += lazy[rt]*(len-len/2); sum[rt<<1|1] += lazy[rt]*(len/2); lazy[rt]=0; } } void build(int l,int r,int rt){ tree[rt].l=l,tree[rt].r=r; lazy[rt]=0; if(l==r){ scanf("%lld",&sum[rt]); return; } int m=tree[rt].mid(); build(lson); build(rson); PushUp(rt); } void update(int d,int l,int r,int rt){ if(tree[rt].l==l&&tree[rt].r==r){ lazy[rt] += d; //打上标记 sum[rt] += d*(r-l+1); return; } if(tree[rt].l==tree[rt].r) return; PushDown(rt,tree[rt].r-tree[rt].l+1);
int m=tree[rt].mid(); if(r<=m) update(d,l,r,rt<<1); else if(l>m) update(d,l,r,rt<<1|1); else{ update(d,l,m,rt<<1); update(d,m+1,r,rt<<1|1); } PushUp(rt); } LL query(int l,int r,int rt){ if(tree[rt].l==l&&tree[rt].r==r){ return sum[rt]; } PushDown(rt,tree[rt].r-tree[rt].l+1); int m = tree[rt].mid(); LL res=0; if(r<=m){ res+=query(l,r,rt<<1); }else if(l>m){ res+=query(l,r,rt<<1|1); }else{ res+=query(l,m,rt<<1); res+=query(m+1,r,rt<<1|1); } return res; } int main(){ int n,m; while(~scanf("%d%d",&n,&m)){ build(1,n,1); while(m--){ char ch[2]; scanf("%s",ch); int a,b,c; if(ch[0]==‘Q‘){ scanf("%d%d",&a,&b); cout<<query(a,b,1)<<endl; }else{ scanf("%d%d%d",&a,&b,&c); update(c,a,b,1); } } } return 0; }
