Problem Description
One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.
1. multiply X with a number.
2. divide X with a number which was multiplied before.
After each operation, please output the number X modulo M.
Input
The first line is an integer T(1≤T≤10),
indicating the number of test cases.
For each test case, the first line are two integers Q and M. Q is the number of
operations and M is described above. (1≤Q≤105,1≤M≤109)
The next Q lines, each line starts with an integer x indicating the type of
operation.
if x is 1, an integer y is given, indicating the number to multiply. (0<y≤109)
if x is 2, an integer n is given. The calculator will divide the number which
is multiplied in the nth operation. (the nth operation must be a type 1
operation.)
It's guaranteed that in type 2 operation, there won't be two same n.
Output
For each test case, the first line, please
output "Case #x:" and x is the id of the test cases starting from 1.
Then Q lines follow, each line please output an answer showed by the
calculator.
Sample Input
1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7
Sample Output
Case #1:
2
1
2
20
10
1
6
42
504
84
题目主要是出现的除法，在模条件下是不能进行除法的，除非存在逆元可以实现除法，但是此处除数不一定与被除数互质。
但是如果过程中不模的话，就要使用大数，会T。
考虑到题目中提到了，除数不会出现相同的。
也就是如果乘了1,2,3,然后再除掉2的话，结果就是由1和3构成，这样就不用考虑每个数的情况了，此时的每个数就是一个整体，结果只和这个数有没有出现有关。
于是可以考虑用线段树来维护分段的积。当某一个数被除掉了，所有与这个数相关的区间都要重新计算，最多有log(q)个区间。
这样效率就是qlogq，是满足条件的。
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <string> #define LL long long using namespace std; const int maxN = 100005; int q, m; int op[maxN], top; //线段树 struct node { int lt, rt; LL val; }tree[4*maxN]; //向上更新 void pushUp(int id) { tree[id].val = (tree[id<<1].val*tree[id<<1|1].val)%m; } //建立线段树 void build(int lt, int rt, int id) { tree[id].lt = lt; tree[id].rt = rt; tree[id].val = 1;//每段的初值，根据题目要求 if (lt == rt) { //tree[id].add = ??; return; } int mid = (lt+rt)>>1; build(lt, mid, id<<1); build(mid+1, rt, id<<1|1); pushUp(id); } void add(int lt, int rt, int id, int pls) { if (lt <= tree[id].lt && rt >= tree[id].rt) { if (pls) { tree[id].val *= pls; tree[id].val %= m; } else tree[id].val = 1; return; } int mid = (tree[id].lt+tree[id].rt)>>1; if (lt <= mid) add(lt, rt, id<<1, pls); if (rt > mid) add(lt, rt, id<<1|1, pls); pushUp(id); } void work() { build(1, q, 1); top = 1; int d, y; for (int i = 0; i < q; ++i) { scanf("%d%d", &d, &y); if (d == 1) add(top, top, 1, y); else add(y, y, 1, 0); op[top++] = y; printf("%I64d ", tree[1].val); } } int main() { //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { printf("Case #%d: ", times); scanf("%d%d", &q, &m); work(); } return 0; }