A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
    | | |
| Case Time Limit: 2000MS | ||
Description
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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
Sample Output
455915
Hint
The sums may exceed the range of 32-bit integers.
Source
, Yang Yi
这个题无非就是更新某个区间然后查询，思路甚至比其它的线段树要简单，但写起来却每次都出错；
思路很简单，如果每次更新都遍历到叶节点，毫无疑问会超时，既然是区间更新，我们可以先将要增加的值存在包含本区间的父亲区间里，下次往子节点操作时再进行类似的操作，也就是加上父亲节点储存的要增加的值，看代码+注释：
#include#include #include #include #include #include using namespace std;const int N=100000+10;int n,m,s[N];struct node{ int l,r; long long n,add;//注意数据范围；} a[N<<2];void build(int l,int r,int k)//构建就没啥好讲的；{ a[k].l=l,a[k].r=r,a[k].add=0; if(l==r) { a[k].n=s[l]; return ; } int mid=(l+r)/2; build(l,mid,2*k); build(mid+1,r,2*k+1); a[k].n=a[k*2].n+a[k*2+1].n;}void update(int l,int r,int c,int k){ if(l<=a[k].l&&a[k].r<=r) { a[k].n+=(a[k].r-a[k].l+1)*c;//满足条件的话，这个区间每个元素都要加上c; a[k].add+=c;将增加的值储存起来，下次更新操作时再往子节点更新； return ; } if(a[k].add)//如果父亲节点增加值还在，那么子节点也要进行更新操作； { a[k*2].add+=a[k].add; a[k*2+1].add+=a[k].add; a[k*2].n+=(a[k*2].r-a[k*2].l+1)*a[k].add; a[k*2+1].n+=(a[k*2+1].r-a[k*2+1].l+1)*a[k].add; a[k].add=0; } int mid=(a[k].l+a[k].r)/2; if(l<=mid) update(l,r,c,2*k); if(r>mid) update(l,r,c,2*k+1); a[k].n=a[k*2].n+a[k*2+1].n;回溯；}long long query(int l,int r,int k){ if(a[k].l==l&&a[k].r==r) return a[k].n; if(a[k].add) { a[k*2].add+=a[k].add; a[k*2+1].add+=a[k].add; a[k*2].n+=(a[k*2].r-a[k*2].l+1)*a[k].add; a[k*2+1].n+=(a[k*2+1].r-a[k*2+1].l+1)*a[k].add; a[k].add=0; } int mid=(a[k].l+a[k].r)/2; if(l>mid) return query(l,r,2*k+1); if(r<=mid) return query(l,r,2*k); return query(l,mid,2*k)+query(mid+1,r,2*k+1);}int main(){ while(~scanf("%d%d",&n,&m)) { memset(a,0,sizeof(a)); for(int i=1; i<=n; i++) scanf("%d",&s[i]); build(1,n,1); while(m--) { int a,b,c; getchar(); char o=getchar(); if(o=='Q') { scanf("%d%d",&a,&b); printf("%I64d\n",query(a,b,1)); } else { scanf("%d%d%d",&a,&b,&c); update(a,b,c,1); } } } return 0;}
其实写出来发现也没什么改变，只不过关键在于更新和查询的时候往子节点所进行的操作；