Function HDU - 5875 （思维）

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博客围绕数组取余查询问题展开，给定含N个正整数的数组A和M个(l,r)形式的查询，定义函数F(l,r)为数组元素取余运算。介绍了输入输出格式，还提到可预处理比自身小的数字位置以跳过部分位置，最后给出求解思路。

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The shorter, the simpler. With this problem, you should be convinced of this truth.

You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate F(l,r), for each query (l,r)

Input

There are multiple test cases.

The first line of input contains a integer T

, indicating number of test cases, and T test cases follow.

  For each test case, the first line contains an integer N(1≤N≤100000).
  The second line contains N space-separated positive integers: A1,…,AN (0≤Ai≤109).
  The third line contains an integer M denoting the number of queries.
  The following M lines each contain two integers l,r (1≤l≤r≤N)

, representing a query.

Output

For each query(l,r)

, output F(l,r)

on one line.

Sample Input

Sample Output

题意：给你一个序列，给出多个询问l，r，求a[l]%a[l+1]%a[l+2]%...%a[r].的结果

由于取余比自己大的数字没有意义，所以只要预处理一下，比自己小的数字的位置，只取余

那些比自己小于等于的数字，所以很多位置可以跳过不计

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100005;
int arr[N],pos[N];
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
			scanf("%d",&arr[i]);
		for(int i=1;i<=n;i++){ //预处理需要计算的位置 
			int tmp=n+1;
			for(int j=i+1;j<=n;j++){
				if(arr[j]<=arr[i]){
					tmp=j;
					break;
				}
			}
			pos[i]=tmp;
		}
		int m;
		scanf("%d",&m);
		while(m--){
			int l,r;
			scanf("%d%d",&l,&r);
			int ans=arr[l];
			for(int i=l+1;i<=r;i=pos[i]) //根据预处理结果进行取模 
				ans%=arr[i];
			printf("%d\n",ans);
		}
	}
	return 0;
}