2015 ACM/ICPC Asia Regional Shanghai Online 1008

1008 An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 5961    Accepted Submission(s): 734


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7

 

[align=left]Sample Output[/align]

Case #1:
2
1
2
20
10
1
6
42
504
84

#include <iostream>
#include <string>
#include <stdlib.h>
#include <ctype.h>
#include <cstdio>
#include <cstdlib>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <math.h>
#include <algorithm>
#define LL long long
#define INF 0x3f3f3f3f
#define RR freopen("in.txt","r",stdin)
#define WW freopen("out.txt","w",stdout)
#define PI acos(-1.0)
using namespace std;
const int maxn = 101010;
bool vis[maxn];
LL h[maxn];
LL ans[maxn];
int main()
{
int T;
int cas = 0;
scanf("%d",&T);
while(T--)
{
memset(vis,false,sizeof(vis));
int x;
LL mod,y,q;
scanf("%lld %lld",&q,&mod);
ans[0] = 1;
printf("Case #%d:\n",++cas);
for(int i=1; i<=q; i++)
{
scanf("%d %lld",&x,&y);
if(x == 1)
{
h[i] = y;
ans[i] =(ans[i-1] * y)%mod;
printf("%I64d\n",ans[i]);
}
else
{
h[i] = y;
vis[h[i]] = true;
vis[i] = true;
for(int j=h[i]; j<=i; j++)
{
ans[j]=ans[j-1];
if(vis[j])
continue;
ans[j] = (ans[j-1] * h[j])%mod;
}
printf("%I64d\n",ans[i]);
}
}
}
return 0;
}