http://www.acmore.net/problem.php?id=1467
根据朴素的欧几里德原理有 gcd(a,b)=gcd(b,a mod b);
则:ax1+by1=bx2+(a mod b)y2;
即:ax1+by1=bx2+(a-[a/b]*b)y2=ay2+bx2-(a/b)*by2;
根据恒等定理得：x1=y2; y1=x2-[a/b]*y2;
本题直接利用这个结论。

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <fstream>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <math.h>
#include <stdlib.h>
using namespace std ;
typedef long long ll;
const int maxn = 100005;
const int mod = 1000000007;
int a[maxn];
ll extend_Eulid(ll a,ll b,ll &x,ll &y){
    if(!b){
        x = 1;
        y = 0;
        return a;
    }
    ll gcd = extend_Eulid(b,a%b,x,y);
    ll temp = x;
    x = y;
    y = temp - a/b*y;
    return gcd;
}
int main(){
    int n;
    while(scanf("%d",&n)&&n){
        ll sum = 1;
        ll x,y;
        for(int i = 0;i < n;i++){
            scanf("%d",a+i);
            sum = (sum*a[i]+mod)%mod;
        }
        extend_Eulid(a[0],mod,x,y);
        printf("%lld",(sum*x%mod+mod)%mod);
        for(int i = 1;i < n;i++){
            extend_Eulid(a[i],mod,x,y);
            printf(" %lld",(sum*x%mod+mod)%mod);
        }
        printf("\n");
    }
    return 0 ;
}