An easy problem
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1697 Accepted Submission(s): 760
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
Source
2015 ACM/ICPC Asia Regional Shanghai Online
既然说是简单题，那就不用想的太复杂，暴力的做法也能过
1 //2016.9.12
2 #include <iostream>
3 #include <cstdio>
4 #include <cstring>
5 #define N 100005
6
7 using namespace std;
8
9 int nu[N], book[N];
10
11 int main()
12 {
13 long long ans;
14 int T, kase = 0, q, mod, op;
15 scanf("%d", &T);
16 while(T--)
17 {
18 ans = 1;
19 memset(book, true, sizeof(book));
20 printf("Case #%d:\n", ++kase);
21 scanf("%d%d", &q, &mod);
22 for(int i = 1; i <= q; i++)
23 {
24 scanf("%d%d", &op, &nu[i]);
25 if(op == 1)
26 {
27 ans *= nu[i];
28 ans %= mod;
29 }
30 else
31 {
32 book[nu[i]] = false;
33 book[i] = false;
34 ans = 1;
35 for(int j = 1; j < i; j++)
36 {
37 if(book[j])ans = (ans*nu[j])%mod;
38 }
39 }
40 printf("%lld\n", ans);
41 }
42 }
43
44 return 0;
45 }