n=6 m =8 x=max(n,m) for i=x to m*n if mod (i,m)=0 and mod (i
r=m MOD n
Mod(m,n)=x x=什么
取模运算,求证(x y) mod m =[(x mod m)(y mod m)] mod m
离散数学题目证明(x·y)(mod m)=((x mod m)·(y mod m))(mod m)
MOD(M,N).EQ.0
a,b,k为大于2的正整数a^k mod (k+1)=n;b^k mod (k+1)=m; 证明 n*m mod (k+
#define N 20 fun(int a[],int n,int m) {int i; for(i=m;i>n;i-
已知a|m=x,a|n=y,a|m+n=z,求i
main() {long m,n,i=1,j,s; scanf("%ld,%ld",&m,&n); for(;i
集合M={x|x=i^n+i^-n,n属于N}中元素个数为几个?
2x+n/m=m/n-(m/n+n/m)x(m+n不等于0)
n mod 2 =