17997 Simple Counting 数学

17997 Simple Counting

时间限制:2000MS  内存限制:65535K
提交次数:0 通过次数:0

题型: 编程题   语言: 不限定

Description

Ly is crazy about counting . Recently , he got a simple problem , but he had to learn Gaoshu these days .So , he turns to you for help .
You are given a sequence A with n positive integers numbered from 1 to n , and then expected to answer Q queries .
Each queries contains an interval [L,R] , you should find the number of index i which satisfies :
                                  {i | Ai mod (i-L+1) = 0 , L <= i <= R }
where Ai mod (i-L+1) = 0 means that Ai can be divided by (i-L+1) .

输入格式

The first line of the input is an integer T , indicates the number of test cases .
Then T cases followed. For each test case :
The first line contains two integers n, Q .
The second line contains n positive integers A1, A2, …. An .
The next Q line , each line contains two integers L, R.

Data range :
1<= T <= 20
1 <= n, Q <= 20000
1<= Ai <= 50000
1<= L <= R <= n

输出格式

For each query, output a single line with an integer indicates the answer expected .

输入样例

2
5 2
1 2 3 4 5
1 5
3 5

6 3
10 7 3 6 24 11
1 3
2 5
5 6

输出样例

5
2
2
3
1

提示

Huge input, scanf is preferred for C/C++.

During first sample ,
for the first query ,A1 mod 1 = 0 , A2 mod 2 = 0 , A3 mod 3 = 0 , A4 mod 4 = 0 ,A5 mod 5 = 0 , so the answer is 5 ;
for the second query , A3 mod 1 = 0 , A4 mod 2 = 0 , A5 mod 3 != 0 , so the answer is 2 .

给定n个数字和m次询问，每次给定区间[L, R]，然后问a[L] % 1 == 0?  a[L + 1] % 2 == 0?，统计答案。

思路：考虑固定左端点L， 就是，对于每一个数字a[i]，首先我们已经知道他在数组里的位置是i。

那么如果k是他的约数的话。如果真有询问问到它是否%k==0？，那么这个时候它应该从哪里开始这段区间的询问呢？

应该要在[i - k + 1, ]处吧，因为i - L + 1 = k。得到L等于这个。

然后就可以用一个vector[L]来保存，以L为开始的区间询问，那些位置会是得到ans的。然后二分小于等于R的个数就可以了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn =  + ;
vector<int>pos[maxn];
void work() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = ; i <= n; ++i) {
        pos[i].clear();
    }
    for (int i = ; i <= n; ++i) {
        int x;
        scanf("%d", &x);
        int end = (int)sqrt(x * 1.0);
        for (int j = ; j <= end; ++j) {
            if (x % j == ) {
                int L = i - j + ;
                if (L >= ) {
                    pos[L].push_back(i);
                } else break;
                if (x / j == j) continue;
                L = i - (x / j) + ;
                if (L >= ) {
                    pos[L].push_back(i);
                }
            }
        }
    }
    for (int i = ; i <= n; ++i) {
        sort(pos[i].begin(), pos[i].end());
    }
    for (int i = ; i <= m; ++i) {
        int L, R;
        scanf("%d%d", &L, &R);
        int ans = upper_bound(pos[L].begin(), pos[L].end(), R) - pos[L].begin();
        printf("%d\n", ans);
    }
}

int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return ;
}