POJ3468 A Simple Problem with Integers （线段树---区间更新）

A Simple Problem with Integers        

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

分析：线段树之区间更新，根据区间更新模板就可以很容易的写出来

AC代码：

#include<cstdio> #include<cstring> using namespace std; typedef long long LL; const int maxn=1e5+10; LL sum[maxn<<2]; LL cnt[maxn<<2]; //延迟标记 int N,Q; void build(int l,int r,int rt){ cnt[rt]=0; if(l==r){ scanf("%lld",&sum[rt]); //输入时注意lld，没注意到WA了一次 return ; } int m=(l+r)>>1; build(l,m,rt<<1); build(m+1,r,rt<<1|1); sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void Down(int rt,int len){ if(cnt[rt]){ cnt[rt<<1]+=cnt[rt]; cnt[rt<<1|1]+=cnt[rt]; sum[rt<<1]+=cnt[rt]*(len-(len>>1)); sum[rt<<1|1]+=cnt[rt]*(len>>1); cnt[rt]=0; } } LL Query(int a,int b,int l,int r,int rt){ if(a<=l && b>=r){ return sum[rt]; } Down(rt,r-l+1); int m=(l+r)>>1; LL ans=0; if(a<=m) ans+=Query(a,b,l,m,rt<<1); if(b>m) ans+=Query(a,b,m+1,r,rt<<1|1); sum[rt]=sum[rt<<1]+sum[rt<<1|1]; return ans; } void Update(int a,int b,int c,int l,int r,int rt){ if(a<=l && b>=r){ cnt[rt]+=c; sum[rt]+=(LL)(r-l+1)*c; return ; } Down(rt,r-l+1); int m=(r+l)>>1; if(a<=m) Update(a,b,c,l,m,rt<<1); if(b>m) Update(a,b,c,m+1,r,rt<<1|1); sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } int main(){ while(scanf("%d%d",&N,&Q)==2){ build(1,N,1); for(int i=0;i<Q;i++){ char ch; int a,b,c; getchar(); scanf("%c",&ch); if(ch=='Q'){ scanf("%d%d",&a,&b); LL ans=Query(a,b,1,N,1); printf("%lld\n",ans); } else { scanf("%d%d%d",&a,&b,&c); Update(a,b,c,1,N,1); } } } return 0; }