Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 2427    Accepted Submission(s): 858

Problem Description

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

1

3

2 3 3

1

1 3

 

 

Sample Output

2

 

 

题目链接：

题意就是给定区间[L,R]，求A[L]%A[L+1]%A[L+2]%……%A[R]的值，这里有一个技巧，可以发现在b>a时，a%b是无意义的，即结果还是a，因此把取模过程改一下，每一次从当前位置idx向右二分找到第一个值小于A[idx]的数，然后对它取模，然后这样迭代地继续寻找，直到在idx~r中找不到这样的位置即可。对了这题其实最大的意义是让我知道了log类函数常数比较大，一开始交用这个函数超时，不管是log2还是log都这样，因此用位运算模拟来求这个指数k。

代码：

#include 
     
      #include 
      
       using namespace std;#define INF 0x3f3f3f3f#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);typedef pair
       
         pii;typedef long long LL;const double PI = acos(-1.0);const int N = 100007;int arr[N], Min[N][18];int Scan(){    int res = 0, ch, flag = 0;    if ((ch = getchar()) == '-')            //判断正负        flag = 1;    else if (ch >= '0' && ch <= '9')          //得到完整的数        res = ch - '0';    while ((ch = getchar()) >= '0' && ch <= '9' )        res = (res << 3) + (res << 1) + ch - '0';    return flag ? -res : res;}void RMQ_init(int l, int r){    int i, j;    for (i = l; i <= r; ++i)        Min[i][0] = arr[i];    for (j = 1; l + (1 << j) - 1 <= r; ++j)        for (i = l; i + (1 << j) - 1 <= r; ++i)            Min[i][j] = min(Min[i][j - 1], Min[i + (1 << (j - 1))][j - 1]);}inline int ST(int l, int r){    int len = r - l + 1;    int k = 0;    while (1 << (k + 1) <= len)        ++k;    return min(Min[l][k], Min[r - (1 << k) + 1][k]);}int getfirstpos(int key, int l, int r){    int L = l, R = r;    int ans = -1;    while (L <= R)    {        int mid = (L + R) >> 1;        if (ST(l, mid) <= key)        {            ans = mid;            R = mid - 1;        }        else            L = mid + 1;    }    return ans;}int main(void){    int tcase, n, m, l, r, i;    scanf("%d", &tcase);    while (tcase--)    {        scanf("%d", &n);        getchar();        for (i = 1; i <= n; ++i)            arr[i] = Scan();        RMQ_init(1, n);        scanf("%d", &m);        while (m--)        {            scanf("%d%d", &l, &r);            if (l == r)                printf("%d\n", arr[l]);            else            {                int idx = l;                int res = arr[l];                int pos;                while (idx + 1 <= r && ~(pos = getfirstpos(res, idx + 1, r)) && res)                {                    res %= arr[pos];                    idx = pos;                }                printf("%d\n", res);            }        }    }    return 0;}