密码学-基于中国剩余定理的门限方案-C++实现

#include
#include
#include
using namespace std;

typedef long long LL;
const int N=100;
LL k=4,n=7,s_start=67187511;
//m1=97,m2=98,m3=99,m4=101,m5=103, 
//m6=107 m7=109 s_start=67187512;;
LL M;
LL m[N],s[N],M_arr[N],N_arr[N],y[N];

void getm(){
	cout<<"请分别输入m1~m"<>m[i]; 
	}
} 


bool if_gcd1(){
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			if(__gcd(m[i],m[j])!=1){
				return false;
			}
		} 
	}
	cout<<"gcd=1满足"<=n-k+2;i--){
		msum2*=m[i];
		
	}
	cout<<"msum1:"<msum2&&msum1>s_start&&msum2