【ZOJ3940 The 13th Zhejiang Provincial Collegiate Programming ContestE】【腦洞 STL-MAP 複雜度分析 區間運算思想 雙指標】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
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
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
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;
}
/*
【題意】
有n（1e5）個數字a[]（1e9），我們有q（1e5）個詢問。
對於每個詢問，想問你——有多少個[0,m]（m∈[0,1e9]）範圍的數，滿足其mod a[1] mod a[2] mod a[3] mod ... mod a[n]== b[i]
（b[]是1e9範圍的數）
【型別】
暴力
複雜度分析
【分析】
這道題的關鍵之處，在於要想到——
取模不僅僅是一個數可以取模，一個區間我們也可以做取模處理。
進一步我們發現，取模得到的區間左界必然都為0
一個區間[0,r)的數 mod a[i]，
如果r>a[i]，那麼——
這個區間會變成r/a[i]個[0,a[i])的區間，以及一個[0,r%a[i])的區間
這樣，我們對於每個a[i]，我們就把所有>a[i]的區間都做處理。
在所有處理都完成之後，我們可以用雙指標的方式處理所有詢問的答案
【時間複雜度&&優化】
O(nlognlogn)
這題的複雜度為什麼是這樣子呢？
對於一個數，這個數做連續若干次的取模運算， 數值變化次數不會超過logn次。
於是我們以區間做取模運算，數值變化次數不會超過nlogn次。
然後加上map的複雜度，總的複雜度不過nlognlogn，可以無壓力AC之。
*/