poj 3468 A Simple Problem with Integers 线段树区间加，区间查询和(模板）
A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://poj.org/problem?id=3468Description
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
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
55
9
15
HINT
题意
区间加，区间查询和
题解：
入门模板题，学了这么久的线段树，竟然打了20分钟，而且还debug了，果然还是太菜了。
忘记建树导致re，忘记return ans。虽然是细节错误，但是说明自己还是不够熟练。什么时候才可以一遍打完，不用编译直接交，不需看结果，帅气切题啊。
代码：
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 using namespace std; 5 #define N 200050 6 #define ll long long 7 int n,m; ll w[N]; 8 struct Tree{int l,r;ll j,sum;}tr[N<<2]; 9 template<typename T>void read(T&x) 10 { 11 int k=0;char c=getchar(); 12 x=0; 13 while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar(); 14 if (c==EOF)exit(0); 15 while(isdigit(c))x=x*10+c-'0',c=getchar(); 16 x=k?-x:x; 17 } 18 void read_char(char &c) 19 {while(!isalpha(c=getchar())&&c!=EOF);} 20 void push_up(int x) 21 { 22 ll len=tr[x].r-tr[x].l+1; 23 tr[x].sum=len*tr[x].j; 24 if (len==1)return; 25 tr[x].sum+=tr[x<<1].sum+tr[x<<1|1].sum; 26 } 27 void push_down(int x) 28 { 29 Tree &a=tr[x<<1],&b=tr[x<<1|1]; 30 a.j+=tr[x].j; 31 b.j+=tr[x].j; 32 push_up(x<<1); 33 push_up(x<<1|1); 34 tr[x].j=0; 35 } 36 void bt(int x,int l,int r) 37 { 38 tr[x].l=l; tr[x].r=r; tr[x].j=0; 39 if (l==r) 40 { 41 tr[x].j=w[l]; 42 push_up(x); 43 return; 44 } 45 int mid=(l+r)>>1; 46 bt(x<<1,l,mid); 47 bt(x<<1|1,mid+1,r); 48 push_up(x); 49 } 50 void update(int x,int l,int r,ll tt) 51 { 52 if (l<=tr[x].l&&tr[x].r<=r) 53 { 54 tr[x].j+=tt; 55 push_up(x); 56 return; 57 } 58 int mid=(tr[x].l+tr[x].r)>>1; 59 if(l<=mid)update(x<<1,l,r,tt); 60 if (mid<r)update(x<<1|1,l,r,tt); 61 push_up(x); 62 } 63 ll query(int x,int l,int r) 64 { 65 if(l<=tr[x].l&&tr[x].r<=r) 66 return tr[x].sum; 67 ll ans=0,mid=(tr[x].l+tr[x].r)>>1; 68 push_down(x); 69 if (l<=mid)ans+=query(x<<1,l,r); 70 if (mid<r)ans+=query(x<<1|1,l,r); 71 push_up(x); 72 return ans; 73 } 74 int main() 75 { 76 #ifndef ONLINE_JUDGE 77 freopen("aa.in","r",stdin); 78 #endif 79 read(n); read(m); 80 for(int i=1;i<=n;i++)read(w[i]); 81 bt(1,1,n); 82 for(int i=1;i<=m;i++) 83 { 84 char id; 85 int x,y; ll tt; 86 read_char(id);read(x);read(y); 87 if (id=='C')read(tt),update(1,x,y,tt); 88 if (id=='Q')printf("%lld\n",query(1,x,y)); 89 } 90 }
