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

 
既然说是简单题,那就不用想的太复杂,暴力的做法也能过
 //2016.9.12
#include <iostream>
#include <cstdio>
#include <cstring>
#define N 100005 using namespace std; int nu[N], book[N]; int main()
{
long long ans;
int T, kase = , q, mod, op;
scanf("%d", &T);
while(T--)
{
ans = ;
memset(book, true, sizeof(book));
printf("Case #%d:\n", ++kase);
scanf("%d%d", &q, &mod);
for(int i = ; i <= q; i++)
{
scanf("%d%d", &op, &nu[i]);
if(op == )
{
ans *= nu[i];
ans %= mod;
}
else
{
book[nu[i]] = false;
book[i] = false;
ans = ;
for(int j = ; j < i; j++)
{
if(book[j])ans = (ans*nu[j])%mod;
}
}
printf("%lld\n", ans);
}
} return ;
}

HDU5475的更多相关文章

  1. hdu-5475 An easy problem---线段树+取模

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5475 题目大意: 给X赋初值1,然后给Q个操作,每个操作对应一个整数M: 如果操作是1则将X乘以对应 ...

  2. HDU5475(线段树)

    An easy problem Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)T ...

  3. ACM学习历程—HDU5475 An easy problem(线段树)(2015上海网赛08题)

    Problem Description One day, a useless calculator was being built by Kuros. Let's assume that number ...

  4. hdu5475(线段树单点修改,统计区间乘积)

    题目意思: 给定a*b*c*d*e*f*....,可以在某一步去掉前面的一个因子,每次回答乘积. #include <cstdio> #include <cstring> #i ...

随机推荐

  1. MC34063中文资料及应用实例(转)

    源:http://blog.chinaunix.net/uid-26199686-id-3207838.html MC34063A(MC33063)芯片器件简介 该器件本身包含了DC/DC变换器所需要 ...

  2. sqlDataAdapter的FillSchema用法

    摘自于网络:http://blog.csdn.net/bupt_zoucq/article/details/6653385 FillSchema是用来向DataTable中填入详细的元数据信息的,例如 ...

  3. 说说Core Animation

    前言 本次分享将从以下方面进行展开: 曾被面试官问倒过的问题:层与视图的关系 CALayer类介绍及层与视图的关系 CAShapeLayer类介绍 UIBezierPath贝塞尔曲线讲解 CoreAn ...

  4. CSS实现三角形方法一--rotate+relative

    方法说明:两个正方形,一个小的,一个大的,将大的正方向进行旋转,然后移动到小的正方形的合适位置,覆盖小正方形的一部分,使小正方形剩余部分为三角形,再把大正方形的背景色改为浏览器窗口的颜色. 用到知识: ...

  5. iOS学习基本常识

    转发至:http://blog.sina.com.cn/s/blog_9266da3d010184i0.html 1. 了解main函数,  UIApplication是初始化程序的核心,它接受4个参 ...

  6. away3d 汽车路线编辑器

    2012年的时候,跟朋友去到一个公司,打算用away3d做一个赛车模拟养成游戏,后来由于种种原因,立项未成,由于朋友已经转行,自己也想对自己做过的事情有一些交代,所以将我负责的部分,赛道编辑器的源码公 ...

  7. SQL复习一(基础知识)

    1.什么是sql SQL(structure query language)是结构化查询语言,它是对关系型数据库的操作语言.它可以应用到所有的数据库中,例如:MySQL.Oracle.SQL serv ...

  8. Arduino单片机使用和开发问题记录(转)

    源:Arduino单片机使用和开发问题记录 1.将程序上传到板子时Arduino IDE提示“avrdude: stk500_getsync(): not in sync: resp=0x00” 网上 ...

  9. ie6,ie7兼容性总结

    摘自: http://www.cnblogs.com/li0803/archive/2009/08/22/1552094.html 其实浏览器的不兼容,我们往往是各个浏览器对于一些标准的定义不一致导致 ...

  10. tableview的reloadData应注意

    http://blog.csdn.net/ouyangtianhan/article/details/7835041 http://stackoverflow.com/questions/160715 ...