HDU 4267 - A Simple Problem with Integers 树状数组区间修改

解法请看http://blog.csdn.net/diannaok/article/details/7959423

挺有收获的。以前虽然会区间修改，但不熟练。没想到可以通过k来解决问题。

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2907    Accepted Submission(s): 938

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 1142 12 22 32 41 2 3 1 22 1 2 22 32 41 1 4 2 12 12 22 32 4

 

Sample Output

111113312341

 

////  4267.cpp//  ACM_HDU////  Created by ipqhjjybj on 13-9-7.//  Copyright (c) 2013年 ipqhjjybj. All rights reserved.//#include<cstdio>#include <iostream>#include <cstring>#define lowbit(x) ((x)&(-x))int a[50050][11][11];int s[50050];int n,q;void Update(int mod,int k,int pos,int add){    while(pos>0){        a[pos][k][mod]+=add;        pos-=lowbit(pos);    }}int Query(int pos,int num){    int i,sum=0;    while(pos<=n){        for(i=1;i<=10;i++)            sum+=a[pos][i][num%i];        pos+=lowbit(pos);    }    return sum;}int main(){    while(scanf("%d",&n)!=EOF){        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)            scanf("%d",s+i);        scanf("%d",&q);        int cas,a,b,k,c;        while(q--){            scanf("%d",&cas);            switch(cas){                case 2:                    scanf("%d",&a);                    printf("%d\n",Query(a,a)+s[a]);                break;                case 1:                    scanf("%d %d %d %d",&a,&b,&k,&c);                    Update(a%k,k,b,c);                    Update(a%k,k,a-1,-c);                break;            }        }    }    return 0;}