ZOJ 3940 Modulo Query(YY+二分)
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.
Where A is an integer array of length N, X is a non-negative integer no greater than M.
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.
题目链接:ZOJ 3940
题意:给出N个数Ai和M,又给Q个询问,每一个询问都是求[0,M]中求是否存在X使得X%A1%A2%A3%......%An=Yi,输出符合有几个这种整数X。
可以发现任何情况下对连续的数取模除非当前值比上一个取模值小,否则直接跳过即可,当然最重要的不是这里,而是如何把题目转换一下,每一次问[0,M]中符合题意的X个数,那么我们可以把[0,M]区间分割为无数个被取模后的小区间,若计这些小区间的贡献均为1,则覆盖在点Yi的情况就是询问Yi的答案,怎么分割呢,当然是用题目给的A数组分割,顺序地输入数组,这里就可以用到上面讲到的取模的技巧来减少分割的次数,分割之后做一遍前缀或后缀和(比如小区间0-1与0-2,0-2显然是包括0-1的,因此是前缀或后缀和关系,两种不同的和只会影响统计时候的加减法问题不会影响答案)。然后询问的时候二分到第一个大于Yi的区间,假设你输入的是3,二分出来的位置是pos,pos对应的子区间为0-4,答案就是从pos~end的贡献和。因为在pos之前只会产生小于3的数,不可能出现3的情况,只有从至少%4开始,才会出现3
代码:
#include <bits/stdc++.h> using namespace std; #define INF 0x3f3f3f3f typedef pair<int, int> pii; typedef long long LL; const int N = 1e5 + 7; const LL MOD = 1e9 + 7; int A[N]; map<int, int>pos; pii prefix[N << 2]; inline int getpos(int l, int r, int key) { int ans = -1; while (l <= r) { int mid = (l + r) >> 1; if (prefix[mid].first > key) { ans = mid; r = mid - 1; } else l = mid + 1; } return ans; } int main(void) { int tcase, n, m, i, q; scanf("%d", &tcase); while (tcase--) { int Min = INF; pos.clear(); scanf("%d%d", &n, &m); pos[m + 1] = 1; prefix[0] = {0, 0}; for (i = 1; i <= n; ++i) { scanf("%d", &A[i]); if (A[i] < Min) Min = A[i]; while (1) { auto it = pos.upper_bound(A[i]); if (it == pos.end()) break; int olde = it->first; int oldcnt = it->second; pos[A[i]] += olde / A[i] * oldcnt; if (olde % A[i] != 0) pos[olde % A[i]] += oldcnt; pos.erase(it); } } auto it = pos.begin(); int sz = 0; while (it != pos.end()) { ++sz; prefix[sz].second = prefix[sz - 1].second + it->second; prefix[sz].first = it->first; ++it; } scanf("%d", &q); LL ans = 0LL; for (i = 1; i <= q; ++i) { LL curans; int x; scanf("%d", &x); if (x >= Min) curans = 0LL; else { int l = getpos(1, sz, x); if (~l) curans = prefix[sz].second - prefix[l - 1].second; else curans = 0LL; } if (curans) { ans = ans + (LL)i * curans % MOD; if (ans > MOD) ans %= MOD; } } printf("%lld\n", ans); } return 0; }