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 ≤ …

链接: https://vjudge.net/problem/POJ-2891 题意: 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…

http://poj.org/problem?id=2891 题意:与中国剩余定理不同,p%ai=bi,此处的ai(i=1 2 3 ……)是不一定互质的,所以要用到的是同余方程组,在网上看到有人称为拓展中国剩余定理. 具体讲解可以看我昨天的博文:http://www.cnblogs.com/KonjakJuruo/p/5176417.html //poj2891 #include<cstdio> #include<cstdlib> #include<cstring> #…

http://poj.org/problem?id=2891 实际上就是一个一元线性同余方程组.按照合并的方式来解即可. 有一个注意点,调用函数是会慢的. #include<iostream> #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; ll t,a1,b1,a2,b2,x_0,y_0; ll ex_gcd(ll a,ll b,ll &x,ll &…

Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 9472   Accepted: 2873 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is…

Strange Way to Express Integers DescriptionElina 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 n…

Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 10907   Accepted: 3336 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is…

F - Strange Way to Express Integers Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers.…

Strange Way to Express Integers Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2891   Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative i…

1635:[例 5]Strange Way to Express Integers sol:貌似就是曹冲养猪的加强版,初看感觉非常没有思路,经过一番艰辛的***,得到以下的结果 随便解释下给以后的自己听:K是要求的数字 第一个读入的A1,Mod1不用改,从2开始做,把Mod2改成LCM,A2改成Ans,接着搞3 /* 原式: X = A[1] (%Mod[1]) X = A[2] (%Mod[2]) ... X = A[n] (%Mod[n]) K[1]*Mod[1]+A[1] = X K[2]…

Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 8370   Accepted: 2508 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is…

题意 Language:Default Strange Way to Express Integers Time Limit: 1000MS Memory Limit: 131072K Total Submissions: 21651 Accepted: 7266 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative inte…

http://poj.org/problem?id=2891 Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 11970   Accepted: 3788 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express no…

Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 8193   Accepted: 2448 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is…

[POJ2891]Strange Way to Express Integers(拓展CRT) 题面 Vjudge 板子题. 题解 拓展\(CRT\)模板题. #include<iostream> #include<cstdio> using namespace std; #define ll long long #define MAX 111111 ll exgcd(ll a,ll b,ll &x,ll &y) { if(!b){x=1,y=0;return a;…

Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 16839   Accepted: 5625 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is…

I. Strange Way to Express Integers 题目描述 原题来自:POJ 2891 给定 2n2n2n 个正整数 a1,a2,⋯,ana_1,a_2,\cdots ,a_na​1​​,a​2​​,⋯,a​n​​ 和 m1,m2,⋯,mnm_1,m_2,\cdots ,m_nm​1​​,m​2​​,⋯,m​n​​,求一个最小的正整数 xxx,满足 ∀i∈[1,n],x≡ai (modmi ),或者给出无解. 输入格式 多组数据. 每组数据第一行一个整数 nnn:接下来 nn…

0.引子 每一个讲中国剩余定理的人,都会从孙子的一道例题讲起 有物不知其数,三三数之剩二,五五数之剩三,七七数之剩二.问物几何? 1.中国剩余定理 引子里的例题实际上是求一个最小的x满足 关键是,其中r1,r2,--,rk互质 这种问题都有多解,每一个解都为最小的解加上若干个lcm(r1,r2,...,rk),这个不用我证了吧(-_-||) 解决这个问题的方法是构造法, 先构造k个数 满足, 这样就保证 ,但是由于 bi 乘了除 ri 以外所有 r,所以bi模其它的 r 都为 0, 再把所有 b…

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 ev…

Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 8476   Accepted: 2554 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…

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 ev…

http://poj.org/problem?id=2891 Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 16849   Accepted: 5630 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is…

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 ≤ …

http://poj.org/problem?id=2891 题意:求最小的$x$使得$x \equiv r_i \pmod{ a_i }$. #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std; typedef long long ll; void ex(ll a, l…

求解方程组 X%m1=r1 X%m2=r2 .... X%mn=rn 首先看下两个式子的情况 X%m1=r1 X%m2=r2 联立可得 m1*x+m2*y=r2-r1 用ex_gcd求得一个特解x' 得到X=x'*m1+r2 X的通解X'=X+k*LCM(m1,m2) 上式可化为:X'%LCM(m1,m2)=X 到此即完成了两个式子的合并,再将此式子与后边的式子合并,最后的得到的X'即为答案的通解,求最小整数解即可. #include<stdio.h> #include<string.h…

http://poj.org/problem?id=2891 (题目链接) 题意 求解线性同余方程组,不保证模数一定两两互质. Solotion 一般模线性方程组的求解,详情请见:中国剩余定理 细节 注意当最后发现方程无解直接退出时,会导致有数据没有读完,然后就会Re,所以先用数组将所有数据存下来. 代码 // poj2891 #include<algorithm> #include<iostream> #include<cstdlib> #include<cst…

题目链接 虽然我不懂... #include <cstdio> #include <cstring> #include <map> #include <cmath> using namespace std; #define LL __int64 LL p[],o[]; LL x,y; LL ext_eulid(LL a,LL b) { LL t,d; ) { x = ; y = ; return a; } d = ext_eulid(b,a%b); t =…

中国剩余定理/扩展欧几里得 题目大意:求一般模线性方程组的解(不满足模数两两互质) solution:对于两个方程 \[ \begin{cases} m \equiv r_1 \pmod {a_1} \\ m \equiv r_2 \pmod{a_2} \end{cases} \] 我们可以列出式子 $$ a_1x+r_1=a_2y+r_2 $$ 利用扩展欧几里得解出一个可行解$M'$.那么我们就可以将两个限制条件合为一个: $$ m \equiv M' \pmod{ lcm(a_1,a_2)}…

求解一元线性同余方程组: x=ri(mod ai) i=1,2,...,k 解一元线性同余方程组的一般步骤:先求出前两个的解,即:x=r1(mod a1)     1x=r2(mod a2)     21式等价于x=r1+a1*m,2式等价于x=r2+a2*n联立可得:m*a1-n*a2=r2-r1=c若方程有解,则必须(a1,a2)|c设d=(a1,a2),那么如果有解,即可求得 m*a1-n*a2=d的解,m=m'则   m*a1-n*a2=c的解,m0=m'*c/d通解m*=m0+(a2/…

题目链接 题意:给k对数,每对ai, ri.求一个最小的m值,令m%ai = ri; 分析:由于ai并不是两两互质的, 所以不能用中国剩余定理. 只能两个两个的求. a1*x+r1=m=a2*y+r2联立得:a1*x-a2*y=r2-r1;设r=r2-r2; 互质的模线性方程组m=r[i](mod a[i]).两个方程可以合并为一个,新的a1为lcm(a1,a2), 新的r为关于当前两个方程的解m,然后再和下一个方程合并…….(r2-r1)不能被gcd(a1,a2)整除时无解.   怎么推出的看…