A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6329 Accepted Submission(s): 2041

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

1 1 1 1

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

题目大意:给予一个数列和两种操作

操作一输入left right mod add

　　　　for i from [left] to [right]

　　　　　　if((i-left)%mod==0) s equence[i]+=add;

操作二输入一下标，返回当前下标值

解题思路: 考虑等式 (i-left)%mod = 0 => i%mod==left%mod

　　　　然后题目转换区间修改和单点查询，使用魔改树状数组解决

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
#define N 50005
int data[N][11][11];
int num[N],n;
int lowbit(int x){
    return x&(-x);
}

void update(int mod,int ans,int cur,int add){///所有 mod， ans(left%mod) 属于一家人 把他们放在一起进行update操作
    while (cur>0){
        data[cur][mod][ans]+=add;
        cur-=lowbit(cur);
    }
}

int query(int x){
    int signs;
    int sum = 0;
    for(int i = 1 ; i <= 10 ; ++i){///查询所有mod的可能情况，得到所有进行过的和
        signs = x;
        while (signs<=n){
            sum+=data[signs][i][x%i];
            signs+=lowbit(signs);
        }
    }
    return sum;
}
int main(){
    int q,ans,left,right,mod,add;
    while (cin>>n){
        memset(data,0, sizeof(data));
        for(int i = 1 ; i <= n ; ++i)scanf("%d",num+i);
        cin>>q;
        while (q--){
            scanf("%d",&ans);
            if(ans==1){
                scanf("%d %d %d %d",&left,&right,&mod,&add);
                update(mod,left%mod,right,add);
                update(mod,left%mod,left-1,-add);
            }
            else{
                scanf("%d",&ans);
                printf("%d\n",query(ans)+num[ans]);
            }
        }
    }
}

HDU 4267 A Simple Problem with Integers

A Simple Problem with Integers

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