首页 > 代码库 > HDU 4267 A Simple Problem with Integers（树状数组区间更新）

HDU 4267 A Simple Problem with Integers（树状数组区间更新）

2024-08-14 02:38:29   222人阅读

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 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
 

 

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 50010using namespace std;int c[12][12][N];//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[N];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;}

 

HDU 4267 A Simple Problem with Integers（树状数组区间更新）


联系
我们