POJ 2891 Strange Way to Express Integers
Description
Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:
Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.
“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”
Since Elina is new to programming, this problem is too difficult for her. Can you help her?
Input
The input contains multiple test cases. Each test cases consists of some lines.
- Line 1: Contains the integer k.
- Lines 2 ~ k + 1: Each contains a pair of integers ai, ri (1 ≤ i ≤ k).
Output
Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.
Sample Input
2
8 7
11 9
Sample Output
31
Hint
All integers in the input and the output are non-negative and can be represented by 64-bit integral types.
Source
POJ Monthly--2006.07.30, Static
题意
给出a_i,r_i
求xequiv r_{i}left( mod a_{i}right)
其中a_i不互质
扩展CRT的应用,算是裸题吧
第一次一遍写对扩欧好感动啊。。。
#include<iostream>
#include<cstdio>
#define LL long long
using namespace std;
const LL MAXN=1e6+10;
LL K,C[MAXN],M[MAXN],x,y;
LL gcd(LL a,LL b)
{
return b==0?a:gcd(b,a%b);
}
LL exgcd(LL a,LL b,LL &x,LL &y)
{
if(b==0){x=1,y=0;return a;}
LL r=exgcd(b,a%b,x,y),tmp;
tmp=x;x=y;y=tmp-(a/b)*y;
return r;
}
LL inv(LL a,LL b)
{
LL r=exgcd(a,b,x,y);
while(x<0) x+=b;
return x;
}
int main()
{
#ifdef WIN32
freopen("a.in","r",stdin);
#else
#endif
while(~scanf("%lld",&K))
{
for(LL i=1;i<=K;i++) scanf("%lld%lld",&M[i],&C[i]);
bool flag=1;
for(LL i=2;i<=K;i++)
{
LL M1=M[i-1],M2=M[i],C2=C[i],C1=C[i-1],T=gcd(M1,M2);
if((C2-C1)%T!=0) {flag=0;break;}
M[i]=(M1*M2)/T;
C[i]= ( inv( M1/T , M2/T ) * (C2-C1)/T ) % (M2/T) * M1 + C1;
C[i]=(C[i]%M[i]+M[i])%M[i];
}
printf("%lldn",flag?C[K]:-1);
}
return 0;
}
- 科技巨头纷纷入局 医疗人工智能需要奋起直追?
- Windows PowerShell 工具
- 游戏开发之在UE4中编写C++代码控制角色
- Visual Studio 64位应用程序编译
- Windows 7 上安装Visual Studio 2015 失败解决方案
- Silverlight调用本机exe程序
- 游戏开发之UE4添加角色到场景中
- 人工智能取代人类?高通副总裁这样说
- Disque:Redis之父新开源的分布式内存作业队列
- mac OS X Yosemite 上编译hadoop 2.6.0/2.7.0及TEZ 0.5.2/0.7.0 注意事项
- EasyStack郭长波连任OpenStack基金会独立董事
- VMware Fusion 中如何复制centos/linux虚拟机
- 浅谈国外航空发动机大数据应用
- asp.net mvc SelectList 的selected 失效及解决方案
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法