Hdu 4267 A Simple Problem with Integers

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原文链接：http://www.cnblogs.com/372465774y/archive/2012/09/12/2682536.html

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本文介绍如何使用线段树解决整数数组上的区间加法和查询问题，包括输入解析、操作类型处理和输出生成。

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1038    Accepted Submission(s): 371

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

Recommend

liuyiding

//怎么说呢、自己真的好弱
//线段树关键在于节点存储信息的意义便于求解
//本来不连续的更新，通过研究、使其成段更新，
h[i][j]代表某个数mod i 的余数为j  ，共有55种情况
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <queue>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
using namespace std;
#define N 50010
struct node
{
    int rec[55];
    int sum;
}st[N<<2];
int h[11][11];
void build(int l,int r,int k)
{
    int m;
    for(m=0;m<55;m++)
     st[k].rec[m]=0;
    if(l==r)
    {
        scanf("%d",&st[k].sum);
        return ;
    }
    m=(l+r)>>1;
    build(lson);
    build(rson);
}
int kk,c,remain;
void update(int &L,int &R,int l,int r,int k)
{
    if(L<=l&&R>=r)
    {
        st[k].rec[h[kk][remain]]+=c;
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m) update(L,R,lson);
    if(R>m)  update(L,R,rson);
}
void down(int &k)
{
    int i;
    for(i=0;i<55;i++)
     {
         if(st[k].rec[i]){
         st[k<<1].rec[i]+=st[k].rec[i];
         st[k<<1|1].rec[i]+=st[k].rec[i];
         st[k].rec[i]=0;
         }
     }
}
int query(int &id,int l,int r,int k)
{
    if(l==r)
    {
        int i,t=0;
        for(i=1;i<=10;i++)
         t+=st[k].rec[h[i][id%i]];
        return t+st[k].sum;
    }
    int m=(l+r)>>1;
    down(k);
    if(id<=m) return query(id,lson);
     else return query(id,rson);
}
int main()
{
    int n;
    int q,i,a,b=0;
    for(i=1;i<=10;i++)
     for(a=0;a<i;a++)
      h[i][a]=b++;
    while(scanf("%d",&n)!=EOF)
    {
        build(1,n,1);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d",&i);
            if(i==1)
            {
                scanf("%d %d %d %d",&a,&b,&kk,&c);
                remain=a%kk;
                update(a,b,1,n,1);
            }
            else
            {
                scanf("%d",&a);
                printf("%d\n",query(a,1,n,1));
            }
        }
    }
    return 0;
}