Hdu 4267 A Simple Problem with Integers

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

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
For each test case, output several lines to answer all query operations.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1 1 1 1 1 3 3 1 2 3 4 1

[align=left]Source[/align]
2012 ACM/ICPC Asia Regional Changchun Online

[align=left]Recommend[/align]
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;
}