题目链接 扩展中国剩余定理:1(直观的).2(详细证明). [Upd:]https://www.luogu.org/problemnew/solution/P4774 #include <cstdio> #include <cctype> #define gc() getchar() typedef long long LL; const int N=1e6+5; LL n,m[N],r[N]; inline LL read() { LL now=0,f=1;register ch…
题目链接 扩展CRT模板题,原理及证明见传送门(引用) #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; ; ll n,m[N],c[N]; void exgcd(ll a,ll b,ll& x,ll& y,ll& g) { ,y=,g=a; else exgcd(b,a%b,y,x,g),y-=x*(a/b); } bool CRT(ll&…
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…
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/…
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-----------------------(…
http://poj.org/problem?id=2891 711323 97935537 475421538 1090116118 2032082 120922929 951016541 15898 418373 161478614 149488440 1748022751 21618619576 810918992 241779667 1772616743 1953316358 125248280 2273149397 3849022001 2509433771 3885219405 35…