Lucky7

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 933    Accepted Submission(s): 345

Problem Description
When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body and sent it back to the shore. It is said that ?? used to surrounded by 7 candles when he faced a extremely difficult problem, and always solve it in seven minutes. 
?? once wrote an autobiography, which mentioned something about himself. In his book, it said seven is his favorite number and he thinks that a number can be divisible by seven can bring him good luck. On the other hand, ?? abhors some other prime numbers and thinks a number x divided by pi which is one of these prime numbers with a given remainder ai will bring him bad luck. In this case, many of his lucky numbers are sullied because they can be divisible by 7 and also has a remainder of ai when it is divided by the prime number pi.
Now give you a pair of x and y, and N pairs of ai and pi, please find out how many numbers between x and y can bring ?? good luck.
 
Input
On the first line there is an integer T(T≤20) representing the number of test cases.
Each test case starts with three integers three intergers n, x, y(0<=n<=15,0<x<y<1018) on a line where n is the number of pirmes. 
Following on n lines each contains two integers pi, ai where pi is the pirme and ?? abhors the numbers have a remainder of ai when they are divided by pi. 
It is guranteed that all the pi are distinct and pi!=7. 
It is also guaranteed that p1*p2*…*pn<=1018 and 0<ai<pi<=105for every i∈(1…n).
 
Output
For each test case, first output "Case #x: ",x=1,2,3...., then output the correct answer on a line.
 
Sample Input
2
2 1 100
3 2
5 3
0 1 100
 
Sample Output
Case #1: 7
Case #2: 14

Hint

For Case 1: 7,21,42,49,70,84,91 are the seven numbers.
For Case2: 7,14,21,28,35,42,49,56,63,70,77,84,91,98 are the fourteen numbers.

好题啊  学到了很多东西...

首先 俄罗斯乘法用于大数取模。中国剩余定理解同模方程组。记住这个解不是唯一的...

容斥原理解决 统计问题

/* ***********************************************
Author :guanjun
Created Time :2016/7/30 13:10:44
File Name :hdu5768.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std; ll a[],m[];
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){
if(!b){d=a;x=1LL;y=0LL;}
else {ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}
}
ll mult(ll a,ll k,ll m){
ll res=;
while(k){
if(k&1LL)res=(res+a)%m;
k>>=;
a=(a<<)%m;
}
return res;
}
ll china(int n,ll *a,ll *m){
ll M=,d,y,x=;
for(int i=;i<n;i++)M*=m[i];
for(int i=;i<n;i++){
ll w=M/m[i];
ex_gcd(m[i],w,d,d,y);
x=(x+mult(y,mult(w,a[i],M),M))%M;
}
return (x+M)%M;
}
ll p[],yu[];
int n;
ll get_ans(ll x){
if(x==)return ;
ll ans=;
int st=(<<n);
for(int i=;i<st;i++){
int cnt=;
ll cur=;
m[cnt]=;a[cnt]=;
cur*=;cnt++;
for(int j=;j<n;j++){
if(i&(<<j)){
m[cnt]=p[j];
a[cnt]=yu[j];
cnt++;
cur*=p[j];
}
}
ll tmp=china(cnt,a,m);
if(tmp>x)continue;
if(cnt&)ans+=(x-tmp)/cur+;
else ans-=(x-tmp)/cur+;
}
//cout<<ans<<endl;
return ans+x/;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int T,t;
ll l,r;
cin>>T;
for(int t=;t<=T;t++){
scanf("%d %I64d %I64d",&n,&l,&r);
for(int i=;i<n;i++)
scanf("%I64d %I64d",&p[i],&yu[i]);
printf("Case #%d: %I64d\n",t,get_ans(r)-get_ans(l-));
}
return ;
}

HDU5768Lucky7的更多相关文章

  1. HDU5768Lucky7(中国剩余定理+容斥定理)(区间个数统计)

    When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortun ...

随机推荐

  1. Hihocoder #1938 最大权闭合子图模板

    这里的讲解很不错,适合作为入坑题: Hihocoder#1938 代码: #include<algorithm> #include<iostream> #include< ...

  2. 在前后端分离的SpringBoot项目中集成Shiro权限框架

    参考[1].在前后端分离的SpringBoot项目中集成Shiro权限框架 参考[2]. Springboot + Vue + shiro 实现前后端分离.权限控制   以及跨域的问题也有涉及

  3. Bootstrap-Table 总结

    Bootstrap-Table 总结 jQuery Java Bootstrap-Table JS文件 传参:直接将需要的参数置于 queryParams 方法中,例如 line:formData注意 ...

  4. 三菱PLC FB库函数调用方法 (Gx Work2版本)

    本文以 GxWorks2 软件为例 1.新建使用标签项目的工程文件 2.从其它库所在工程项目中导入库 3.选择库文件及FB功能块 4.插入FB功能块调用

  5. 【04】 CSS开发注意事项

    [04] CSS注意事项 1. 页面编码规范 1.1. 统一使用 UTF-8 编码,用@charset "utf-8"指定页面编码. 1.2. 全局字体设置: windows 7系 ...

  6. 【05】emmet系列之各种缩写

    [01]emmet系列之基础介绍 [02]emmet系列之HTML语法 [03]emmet系列之CSS语法 [04]emmet系列之编辑器 [05]emmet系列之各种缩写 各种缩写   缩写:! & ...

  7. 用java实现二分搜索<算法分析>

    实验目的:1.复习java编程:2.掌握二分搜索技术的基本原理:3.掌握使用java程序进行二分搜索的方法.实验步骤:1.由用户输入5个以上的整数:2.利用二分搜索算法完成对数组的搜索. packag ...

  8. hdu 2433 Travel (最短路树)

     One day, Tom traveled to a country named BGM. BGM is a small country, but there are N (N <= 100) ...

  9. 桐桐的糖果计划(vijos 1325)

    背景 桐桐是一个快乐的小朋友,他生活中有许多许多好玩的事,让我们一起来看看吧…… 描述 桐桐很喜欢吃棒棒糖.他家处在一大堆糖果店的附近. 但是,他们家的区域经常出现塞车.塞人等情况,这导致他不得不等到 ...

  10. Pull方式解析XML文件

    package com.pingyijinren.test; import android.content.Intent; import android.os.Handler; import andr ...