A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 130735 | Accepted: 40585 | |
| 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 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.
分析：要求是在[l,r]的区间内修改或者查询和，当然会想到树状数组。但是因为每次是区间修改，所以需要转化一下。
首先，假设这里我们只考虑单点查询。新建一个数组b[i]来存储每次的修改信息。对于每一个C l r d，将b[l]加上d，在将b[r+1]减去d，那么每次查询的时候就输出a[x]+（b[x]的前缀和）就可以得到a[x]修改后的值。正确性易证，画图就很好理解了，这里蒟蒻就不画图了（偷懒一波）。
那么再考虑区间查询，易得a[1~x]整体修改的值为Σxi=1Σij=1b[j]，推导Σxi=1Σij=1b[j]=Σxi=1(x-i+1)*b[i]=(x+1)Σxi=1b[i]-Σxi=1i*b[i](格式不太好看将就下吧)。那么这题的算法就可以确定了。
建立两个树状数组c0,c1，对于每一个修改操作，执行以下操作：
将c0中的l位置加d，将c0中的r+1位置减d
将c1中的l位置加l*d，将c1中的r+1位置减（r+1）*d
再用sum[]直接记录a[]的前缀和，对于每一个询问指令，输出(sum[r]+(r+1)*get(c0,r)-get(c1,r))-(sum[l-1]+l*get(c0,l-1)-get(c1,l-1))。实际上也就是用的一般的前缀和与树状数组相结合，并且运用了差分的思想。
Code：
1 //It is made by HolseLee on 17th May 2018
2 //POJ 3468
3 #include<cstdio>
4 #include<cstring>
5 #include<cstdlib>
6 #include<cmath>
7 #include<iostream>
8 #include<iomanip>
9 #include<algorithm>
10 #define Fi(i,a,b) for(int i=a;i<=b;i++)
11 using namespace std;
12 typedef long long ll;
13 const int N=1e5+7;
14 ll n,m,sum[N],c[2][N],ans;
15 inline ll lowbit(int x){return x&-x;}
16 inline void add(int k,int x,int y)
17 {for(int i=x;i<=n;i+=lowbit(i))c[k][i]+=y;}
18 inline ll get(int k,int x)
19 {ll ret=0;for(int i=x;i>=1;i-=lowbit(i))ret+=c[k][i];return ret;}
20 int main()
21 {
22 ios::sync_with_stdio(false);
23 cin>>n>>m;int x,y,z;char opt;
24 Fi(i,1,n)cin>>x,sum[i]=sum[i-1]+x;
25 Fi(i,1,m){cin>>opt;
26 if(opt==‘C‘){cin>>x>>y>>z;
27 add(0,x,z);add(0,y+1,-z);
28 add(1,x,x*z);add(1,y+1,-(y+1)*z);}
29 else {cin>>x>>y;
30 ll ans=(sum[y]+(y+1)*get(0,y)-get(1,y));
31 ans-=(sum[x-1]+x*get(0,x-1)-get(1,x-1));
32 printf("%lld\n",ans);}}
33 return 0;
34 }
原文地址：https://www.cnblogs.com/cytus/p/9052893.html