3212: Pku3468 A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 810 Solved: 354
[Submit][Status]
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.
Source
题解：
想了想树状数组如何维护区间修改，区间求和，觉得貌似很简单？
差分了之后 用树状数组维护 a[i] 以及 i*a[i] 即可
sum(1,n)=(n+1)*sigma(a[i])-sigma(i*a[i])
靠着树状数组刷进了第一页。。。
代码：
1 #include<cstdio>
2
3 #include<cstdlib>
4
5 #include<cmath>
6
7 #include<cstring>
8
9 #include<algorithm>
10
11 #include<iostream>
12
13 #include<vector>
14
15 #include<map>
16
17 #include<set>
18
19 #include<queue>
20
21 #include<string>
22
23 #define inf 1000000000
24
25 #define maxn 100000+1000
26
27 #define maxm 500+100
28
29 #define eps 1e-10
30
31 #define ll long long
32
33 #define pa pair<int,int>
34
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42
43 #define mod 1000000007
44
45 using namespace std;
46
47 inline ll read()
48
49 {
50
51 ll x=0,f=1;char ch=getchar();
52
53 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
54
55 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
56
57 return x*f;
58
59 }
60 ll s[2][maxn],n,m;
61 inline void add(int k,int x,ll y)
62 {
63 for(;x<=n;x+=x&(-x))s[k][x]+=y;
64 }
65 inline ll sum(int k,int x)
66 {
67 ll t=0;
68 for(;x;x-=x&(-x))t+=s[k][x];
69 return t;
70 }
71
72 int main()
73
74 {
75
76 freopen("input.txt","r",stdin);
77
78 freopen("output.txt","w",stdout);
79
80 n=read();m=read();
81 ll x=0,y,z;
82 for(ll i=1;i<=n;i++)
83 {
84 y=read();
85 add(0,i,y-x);
86 add(1,i,i*(y-x));
87 x=y;
88 }
89 while(m--)
90 {
91 char ch[2];
92 scanf("%s",ch);
93 if(ch[0]==‘C‘)
94 {
95 x=read();y=read();z=read();
96 add(0,x,z);add(1,x,x*z);
97 add(0,y+1,-z);add(1,y+1,(-z)*(y+1));
98 }
99 else
100 {
101 x=read();y=read();
102 ll sum1=(x)*sum(0,x-1)-sum(1,x-1);
103 ll sum2=(y+1)*sum(0,y)-sum(1,y);
104 printf("%lld\n",sum2-sum1);
105 }
106 }
107
108 return 0;
109
110 }