HDOJ 4267 A Simple Problem with Integers (线段树）

题目：

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

思路：

区间修改单点查询

第一个操作是对在a-b区间内的位置i 如果满足(i-a)%k==0 就把这个位置上的值加上c

式子可以等同于i%k==a%k 所以问题就转化为了右边的部分

从数据范围中可以看到k的范围很小对k进行枚举

k=1时可以取的余数为0，1

k=2时可以取的余数为0，1 ，2

以此类推所有可以取的结果共55种根据取余的情况对这一个区间内该更新哪些线段树进行纪录然后对这一棵线段树这个区间内的所有值进行更新

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int maxn=5e4+10;
int n,m,y,a,b,k,c,op,ans;
int x[maxn];

struct node{
    int l,r,w;
    int add[55];
}tree[maxn<<2];

void build(int l,int r,int rt){
    tree[rt].l=l;
    tree[rt].r=r;
    memset(tree[rt].add,0,sizeof(tree[rt].add));
    if(l==r){
        tree[rt].w=x[l];
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,rt*2);
    build(mid+1,r,rt*2+1);
    tree[rt].w=tree[rt*2].w+tree[rt*2+1].w;
}

void pushdown(int rt){
    tree[rt*2].w+=tree[rt].w;
    tree[rt*2+1].w+=tree[rt].w;
    tree[rt].w=0;
    for(int i=0;i<55;i++){
        tree[rt*2].add[i]+=tree[rt].add[i];
        tree[rt*2+1].add[i]+=tree[rt].add[i];
        tree[rt].add[i]=0;
    }
}

void update(int tmp,int rt){
    if(tree[rt].l>=a && tree[rt].r<=b){
        int index=k*(k-1)/2+tmp;
        tree[rt].add[index]+=c;
        tree[rt].w+=c;
        return;
    }
    if(tree[rt].w) pushdown(rt);
    int mid=(tree[rt].l+tree[rt].r)/2;
    if(a<=mid) update(tmp,rt*2);
    if(b>mid) update(tmp,rt*2+1);
//    tree[rt].w=tree[rt*2].w+tree[rt*2+1].w;
}

void query(int rt){
    if(tree[rt].l==y && tree[rt].r==y){
        for(int i=1;i<=10;i++){
            int index=i*(i-1)/2+y%i;
            ans+=tree[rt].add[index];
        }
        return;
    }
    if(tree[rt].w) pushdown(rt);
    int mid=(tree[rt].l+tree[rt].r)/2;
    if(y<=mid) query(rt*2);
    else query(rt*2+1);
}

int main(){
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++){
            scanf("%d",&x[i]);
        }
        build(1,n,1);
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            scanf("%d",&op);
            if(op==1){
                scanf("%d%d%d%d",&a,&b,&k,&c);
                update(a%k,1);
            }
            if(op==2){
                ans=0;
                scanf("%d",&y);
                query(1);
                printf("%d\n",x[y]+ans);
            }
        }
    }
    return 0;
}

HDOJ 4267 A Simple Problem with Integers (线段树）

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