http://poj.org/problem?id=2891 结果看了半天还是没懂那个模的含义...懂了我再补充... 其他的思路都在注释里 /********************* Template ************************/ #include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include <deq…
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 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: 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 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…
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…
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 题意:求解一个数x使得 x%8 = 7,x%11 = 9; 若x存在,输出最小整数解.否则输出-1: ps: 思路:这不是简单的中国剩余定理问题,由于输入的ai不一定两两互质,而中国剩余定理的条件是除数两两互质. 这是一般的模线性方程组,对于 X mod m1=r1 X mod m2=r2 ... ... ... X mod mn=rn 首先,我们看两个式子的情况 X mod m1=r1-----------------------(…
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…
一种不断迭代,求新的求余方程的方法运用中国剩余定理. 总的来说,假设对方程操作.和这个定理的数学思想运用的不多的话.是非常困难的. 參照了这个博客的程序写的: http://scturtle.is-programmer.com/posts/19363.html 这个博客举例说的挺好的:http://blog.csdn.net/mishifangxiangdefeng/article/details/7109217 hdu 3579 Hello Kiki 中国剩余定理(不互质的情况) 对互质的情况…