连续取模-function

2017-09-22 21:56:08

The shorter, the simpler. With this problem, you should be convinced of this truth. 
   
  You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as: 
F(l,r)={AlF(l,r−1) modArl=r;l<r.F(l,r)={All=r;F(l,r−1) modArl<r. 
You job is to calculate F(l,r)F(l,r), for each query (l,r)(l,r).

InputThere are multiple test cases. 
   
  The first line of input contains a integer TT, indicating number of test cases, and TT test cases follow. 
   
  For each test case, the first line contains an integer N(1≤N≤100000)N(1≤N≤100000). 
  The second line contains NN space-separated positive integers: A1,…,AN (0≤Ai≤109)A1,…,AN (0≤Ai≤109). 
  The third line contains an integer MM denoting the number of queries. 
  The following MM lines each contain two integers l,r (1≤l≤r≤N)l,r (1≤l≤r≤N), representing a query.

OutputFor each query(l,r)(l,r), output F(l,r)F(l,r) on one line.Sample Input

1
3
2 3 3
1
1 3

Sample Output

2

代码如下：

#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>

using namespace std;

#define MAXN 100010

int a[MAXN],nex[MAXN];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,j,n,m;
        scanf("%d",&n);
        for(i = ; i<=n; ++i)
        {
            scanf("%d",&a[i]);
        }
        for(i = ;i<=n;++i)
        {
            nex[i] = -;
            for(j = i+;j<=n;++j)
            {
                if(a[j]<=a[i])
                {
                    nex[i] = j;
                    break;
                }
            }
        }
        scanf("%d",&m);
        for(i = ;i<m;++i)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int num = a[l];
            for(j = nex[l];j<=r;j = nex[j])
            {
                if(j == -)
                {
                    break;
                }
                num%=a[j];
            }
            printf("%d\n",num);
        }
    }
    return ;
}