集训队专题（10）1002 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): 4985    Accepted Submission(s): 1569

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

 

此题为区间更新，区间查询，但是在区间更新上，会有点要求，注意点就好

#include <cstdio>#include <cstring>#include <algorithm>#define LL __int64using namespace std;const int maxn = 50010;struct tree{    int l , r, k[12];}a[4*maxn];int num[maxn] , N , Q;void build(int l , int r , int k){    a[k].l = l;    a[k].r = r;    for(int i = 0; i < 12; i++) a[k].k[i] = 0;    if(l != r){        int mid = (l+r)/2;        build(l , mid , 2*k);        build(mid+1 , r , 2*k+1);    }}void update(int l , int r , int k , int lk , int lc){    if(l <= a[k].l && a[k].r <= r) a[k].k[lk] += lc;    else{        int mid = (a[k].l+a[k].r)/2;        if(mid >= r) update(l , r, 2*k , lk , lc);        else if(mid < l) update(l , r , 2*k+1 , lk , lc);        else{            update(l , mid , 2*k , lk , lc);            int tl = mid+1+(lk-(mid+1-l)%lk)%lk;            if(tl <= r) update(tl , r , 2*k+1 , lk , lc);        }    }}int getAns(int k , int i){    int ans = 0;    for(int j = 1; j <= 10; j++){        if((i-a[k].l)%j == 0) ans += a[k].k[j];    }    return ans;}int query(int l , int r , int k){    if(l <= a[k].l && a[k].r <= r) return getAns(k , l);    else{        int mid = (a[k].l+a[k].r)/2;        int ans = getAns(k , l);        if(mid >= r) return ans+query(l , r , 2*k);        else return ans+query(l , r , 2*k+1);    }}int main(){    while(~scanf("%d" , &N)){        for(int i = 1; i <= N; i++) scanf("%d" , &num[i]);    build(1 , N , 1);    scanf("%d" , &Q);    int op , la, lb , lk , lc;    while(Q--){        scanf("%d" , &op);        if(op == 1){            scanf("%d%d%d%d" , &la , &lb , &lk , &lc);            update(la , lb , 1 , lk , lc);        }else{            scanf("%d" , &la);            printf("%d\n" , num[la]+query(la , la , 1));        }    }    }    return 0;}