A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4032    Accepted Submission(s): 1255

Problem Description

Let
A1, A2, ... , AN be N elements. You need to deal with two kinds of
operations. One type of operation is to add a given number to a few
numbers in a given interval. The other is to query the value of some
element.

Input

There are a lot of test cases.
The first line contains an integer N. (1 <= N <= 50000)
The
second line contains N numbers which are the initial values of A1, A2,
... , AN. (-10,000,000 <= the initial value of Ai <= 10,000,000)
The third line contains an integer Q. (1 <= Q <= 50000)
Each of the following Q lines represents an operation.
"1
a b k c" means adding c to each of Ai which satisfies a <= i <= b
and (i - a) % k == 0. (1 <= a <= b <= N, 1 <= k <= 10,
-1,000 <= c <= 1,000)
"2 a" means querying the value of Aa. (1 <= a <= N)

Output

For each test case, output several lines to answer all query operations.

Sample Input

4
1 1 1 1
14
2 1
2 2
2 3
2 4
1 2 3 1 2
2 1
2 2
2 3
2 4
1 1 4 2 1
2 1
2 2
2 3
2 4

Sample Output

1
1
1
1
1
3
3
1
2
3
4
1

(i-a)%k==0   即  i%k==a%k             分组   x%k==a%k的为一组，   参数 mod, k,x

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <set>
 7 #include <map>
 8 #include <string>
 9 #include <vector>
10 #include <queue>
11 #include <stack>
12 #include <iomanip>
13 #include <sstream>
14 using namespace std;
15 //#define local
16 typedef long long LL;
17 const int INF=0x4fffffff;
18 const int EXP=1e-5;
19 const int MS=50005;
20
21 int C[11][11][MS];      //C[mod][k][x]
22 int num[MS];
23
24 int lowbit(int x)
25 {
26       return x&(-x);
27 }
28
29 //   修改区间，单点求职，   树状数组需要逆过来。
30
31 void updata(int mod,int k,int x,int d)
32 {
33       while(x>0)
34       {
35             C[mod][k][x]+=d;
36             x-=lowbit(x);
37       }
38 }
39
40 int getsum(int a,int x)
41 {
42       int res=0;
43       while(x<MS)     //x<=n
44       {
45             for(int k=1;k<=10;k++)
46             {
47                   res+=C[a%k][k][x];
48             }
49             x+=lowbit(x);
50       }
51       return res;
52 }
53
54 int main()
55 {
56       #ifdef local
57       freopen("in.txt","r",stdin);
58       freopen("out.txt","w",stdout);
59       #endif // local
60       int n;
61       while(scanf("%d",&n)!=EOF)
62       {
63             for(int i=1;i<=n;i++)
64                   scanf("%d",&num[i]);
65             memset(C,0,sizeof(C));
66             int m;
67             scanf("%d",&m);
68             while(m--)
69             {
70                   int op,a,b,k,c;
71                   scanf("%d",&op);
72                   if(op==1)
73                   {
74                         scanf("%d%d%d%d",&a,&b,&k,&c);
75                         updata(a%k,k,b,c);
76                         updata(a%k,k,a-1,-c);
77                   }
78                   else
79                   {
80                         scanf("%d",&a);
81                         int ans=getsum(a,a);
82                         printf("%d\n",ans+num[a]);
83                   }
84             }
85       }
86       return 0;
87 }