BZOJ3212: Pku3468 A Simple Problem with Integers

3212: Pku3468 A Simple Problem with Integers

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 810  Solved: 354
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Description

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 }
View Code

 

posted @ 2014-09-28 12:46  ZYF-ZYF  Views(216)  Comments(0Edit  收藏  举报