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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 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
题解：线段树延迟做法，就是延迟对孩子节点的操作，把前面要操作的数据记录下来，等下次必须访问孩子的时候才去做，减少递归次数。
#include <iostream>
#include <cstdio>
#include <cstring>
#define mem(a) memset(a,0,1izeof(a));
using namespace std;
long long arr[400005];
long long add[400005];
long long a[100005];
void pushUp(int k)
{
arr[k] = arr[k << 1] + arr[(k << 1) | 1];
}
void pushDown(int k,int x) //分给左右孩子。
{
if(add[k] != 0)
{
add[k << 1] += add[k];
add[(k << 1) | 1] += add[k];
arr[k << 1] += add[k] * (x - (x >> 1));
arr[(k << 1) | 1] += add[k] * (x >> 1);
add[k] = 0;
}
}
void segTree(int k,int l,int r)
{
add[k] = 0;
if(l == r)
{
arr[k] = a[l];
return;
}
int mid = ( l + r) >> 1;
segTree(k << 1,l,mid);
segTree((k << 1) | 1,mid + 1,r);
pushUp(k);
}
void update(int k,int l,int r,int x,int y,int c)
{
if(l == x && r == y)
{
add[k] += c;
arr[k] += (r - l + 1) * c;
return;
}
int mid = (l + r) >> 1;
pushDown(k,r - l + 1);
if(x > mid)
{
update((k << 1) | 1,mid + 1,r,x,y,c);
}
else if(y <= mid)
{
update(k << 1,l,mid,x,y,c);
}
else
{
update(k << 1,l,mid,x,mid,c);
update((k << 1) | 1,mid + 1,r,mid + 1,y,c);
}
pushUp(k);
}
long long query(int k,int l,int r,int x,int y)
{
if(l == x && r == y)
{
return arr[k];
}
pushDown(k,r - l + 1);
int mid = (l + r) >> 1;
if(x > mid)
{
return query((k << 1) | 1,mid + 1,r,x,y);
}
if(y <= mid)
{
return query(k << 1,l,mid,x,y);
}
return query(k << 1,l,mid,x,mid) + query((k << 1) | 1,mid + 1,r,mid + 1,y);
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m) != EOF)
{
for(int i = 1;i <= n;i++)
{
scanf("%lld",a + i);
}
segTree(1,1,n);
char s[3];
int l,r;
for(int i = 0;i < m;i++)
{
scanf("%s%d%d",s,&l,&r);
if('Q' == s[0])
{
printf("%lld\n",query(1,1,n,l,r));
}
else
{
long long c;
scanf("%lld",&c);
update(1,1,n,l,r,c);
}
}
}
return 0;
}
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A Simple Problem with Integers
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原文地址：http://blog.csdn.net/wang2534499/article/details/47376713
