Function HDU - 5875(暴力+思维)_function 5875-优快云博客

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/82260664

博客围绕HDU - 5875的Function问题展开，给定含N个正整数的数组A及M个(l,r)形式的查询，定义了函数F(l,r)。需根据输入的N、数组A、查询数M及具体查询(l,r)输出F(l,r)。分析指出预处理右边首个比它小的数，询问时跳着向后找，可直接暴力求解。

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Function HDU - 5875

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) = {A l l = r; F (l, r - 1) m o d A r l < r

$F(l,r) = \begin{cases} \ A_l\ \ \ \ \ \ \ l = r;\\ \ F(l,r-1)\ mod\ A_r\ \ \ \ \ \ \ \ l < r \end{cases}$
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：

给定区间l，r求上述公式

分析：

预处理出右边第一个比它小的数，然后询问的话就可以跳着向后找，直接暴力就可以

code：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000+7;
int pos[maxn];
int a[maxn],n,m;
void init(){
    memset(pos,-1,sizeof(pos));
    for(int i = n-1; i >= 1; i--){
        int p = i + 1;
        for(;;){
            if(a[i] > a[p]){
                pos[i] = p;
                break;
            }
            if(pos[p] == -1){
                pos[i] = -1;
                break;
            }
            p = pos[p];
        }
    }
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i]);
        }
        init();
        scanf("%d",&m);
        for(int i = 1; i <= m; i++){
            int l,r;
            scanf("%d%d",&l,&r);
            int ans = a[l];
            l = pos[l];
            while(l <= r && l != -1){
                ans %= a[l];
                l = pos[l];
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}