A Simple Problem with Integers

                                  Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

Source

2012 ACM/ICPC Asia Regional Changchun Online

题意：

题解：树状数组的区间修改，单点查询，注意这里的不连续区间修改

///1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a) scanf("%d",&a)
#define g 9.8
#define mod 100000000
#define eps 0.0000001
#define maxn 50000+1
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)
    {
        if(ch==‘-‘)f=-1;
        ch=getchar();
    }
    while(ch>=‘0‘&&ch<=‘9‘)
    {
        x=x*10+ch-‘0‘;
        ch=getchar();
    }
    return x*f;
}
//****************************************

int c[maxn][11][11],s[maxn],n;
int lowbit(int x){return x&(-x);}
void update(int a,int b,int x,int v)
{
    while(x<=n)
    {
        c[x][a][b]+=v;
        x+=lowbit(x);
    }
}
int get(int x,int a)
{
    int sum=0;
    while(x>0)
    {
        FOR(i,1,10)
        sum+=c[x][i][a%i];
        x-=lowbit(x);
    }
    return sum;
}
int main()
{

    while(READ(n)!=EOF)
    {
        mem(c);
        FOR(i,1,n)
        {
            scanf("%d",&s[i]);
        }
        int q=read();
        int t,a,b,k,add;
        FOR(i,1,q)
        {
            scanf("%d",&t);
            if(t==1)
            {
                scanf("%d%d%d%d",&a,&b,&k,&add);
                update(k,a%k,a,add);
                update(k,a%k,b+1,-add);
            }
            else {
                    scanf("%d",&a);
                printf("%d\n",get(a,a)+s[a]);
            }
       }
    }
    return 0;
}

代码