A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5402    Accepted Submission(s): 1710

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

/*
类似的区间更新，但这个区间不是实际意义上的区间而是，i到j满足条件的点更新
这个题和区间更新类似，用另一个数组维护，满足条件的点的前缀和，询问的时候直接用原数组的值加上满足条件的值
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#define lowbit(x) x&(-x)
#define N 50010
using namespace std;
int c[12][12]
;//
int n,q,op;
void update(int s1,int s2,int x,int val)//间隔为s1,起点为s2,需要更新的点为x
{
while(x<=n)
{
c[s1][s2][x]+=val;
x+=lowbit(x);
}
}
int getsum(int s1,int s2,int x)
{
int s=0;
while(x>0)
{
s+=c[s1][s2][x];
x-=lowbit(x);
}
return s;
}
int num
;
int main()
{
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
while(scanf("%d",&n)!=EOF)
{
//cout<<n<<endl;
memset(c,0,sizeof c);
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
scanf("%d",&q);
int x,y,k,val;
while(q--)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d%d%d",&x,&y,&k,&val);
x--;
y--;

int knum=(y-x)/k;//需要更新的点的个数
int s1=x%k;//这次更新的起点
update(k,s1,x/k+1,val);
update(k,s1,x/k+knum+2,-val);
}
else if(op==2)
{
scanf("%d",&x);
x--;
int cur=num[x];
for(int i=1;i<=10;i++)//遍历的是k的取值
cur+=getsum(i,x%i,x/i+1);
printf("%d\n",cur);
}
}
}
return 0;
}