【ZOJ3940 The 13th Zhejiang Provincial Collegiate Programming ContestE】【脑洞 STL-MAP 复杂度分析 区间运算思想 双指针】M
One day, Peter came across a function which looks like:
- F(1, X) = X mod A1.
- F(i, X) = F(i - 1, X) mod Ai, 2 ≤ i ≤ N.
Peter wants to know the number of solutions for equation F(N, X) = Y, where Y is a given number.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers N and M (2 ≤ N ≤ 105, 0 ≤ M ≤ 109).
The second line contains N integers: A1, A2, ..., AN (1 ≤ Ai ≤ 109).
The third line contains an integer Q (1 ≤ Q ≤ 105) - the number of queries. Each of the following Q lines contains an integer Yi (0 ≤ Yi ≤ 109), which means Peter wants to know the number of solutions for equation F(N, X) = Yi.
Output
For each test cases, output an integer S = (1 ⋅ Z1 + 2 ⋅ Z2 + ... + Q ⋅ ZQ) mod (109 + 7), where Zi is the answer for the i-th query.
Sample Input
1 3 5 3 2 4 5 0 1 2 3 4
Sample Output
8
Hint
The answer for each query is: 4, 2, 0, 0, 0.
Author: LIN, XiSource: The 13th Zhejiang Provincial Collegiate Programming Contest
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int casenum, casei;
int n, m, q;
map<int, int>mop;
map<int, int>::iterator it;
pair<int, int>a[N];
int main()
{
scanf("%d", &casenum);
for (casei = 1; casei <= casenum; ++casei)
{
mop.clear();
scanf("%d%d", &n, &m);
mop[m + 1] = 1;
for (int i = 1; i <= n; ++i)
{
int x; scanf("%d", &x);
while(1)
{
it = mop.upper_bound(x);
if (it == mop.end())break;
mop[x] += it->first / x * it->second;
if(it->first%x)mop[it->first%x] += it->second;
mop.erase(it);
}
}
int sum = 0;
for (it = mop.begin(); it != mop.end(); ++it)sum += it->second;
scanf("%d", &q);
for (int i = 1; i <= q; ++i)scanf("%d", &a[i].first), a[i].second = i;
sort(a + 1, a + q + 1);
it = mop.begin();
int ans = 0;
for (int i = 1; i <= q; ++i)
{
while (a[i].first >= it->first)
{
sum -= it->second;
++it;
if (it == mop.end())break;
}
if (it == mop.end())break;
ans = (ans + (LL)sum * a[i].second) % Z;
}
printf("%d\n", ans);
}
return 0;
}
/*
【题意】
有n（1e5）个数字a[]（1e9），我们有q（1e5）个询问。
对于每个询问，想问你——有多少个[0,m]（m∈[0,1e9]）范围的数，满足其mod a[1] mod a[2] mod a[3] mod ... mod a[n]== b[i]
（b[]是1e9范围的数）
【类型】
暴力
复杂度分析
【分析】
这道题的关键之处，在于要想到——
取模不仅仅是一个数可以取模，一个区间我们也可以做取模处理。
进一步我们发现，取模得到的区间左界必然都为0
一个区间[0,r)的数 mod a[i]，
如果r>a[i]，那么——
这个区间会变成r/a[i]个[0,a[i])的区间，以及一个[0,r%a[i])的区间
这样，我们对于每个a[i]，我们就把所有>a[i]的区间都做处理。
在所有处理都完成之后，我们可以用双指针的方式处理所有询问的答案
【时间复杂度&&优化】
O(nlognlogn)
这题的复杂度为什么是这样子呢？
对于一个数，这个数做连续若干次的取模运算， 数值变化次数不会超过logn次。
于是我们以区间做取模运算，数值变化次数不会超过nlogn次。
然后加上map的复杂度，总的复杂度不过nlognlogn，可以无压力AC之。
*/