题意;给出N,A,B;求A*x+ B*y = N+1 的大于0 的解的数量;
思路:先用exgcd求出大于0的初始解x,rest = N - x*A; sum = rest/LCM(A, B);
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <time.h>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set> #define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 100005
const double PI = acos(-1.0);
typedef long long LL ;
LL x, y, A, B, n, C;
LL gcd(LL a, LL b){if(a== ) return b;else return gcd(b%a,a);} LL exgcd(LL a, LL b, LL &x, LL &y){
if(b == ){
x = ; y = ;return a;
}else{
LL t = exgcd(b, a%b, y, x);
y -= (a/b)*x;
return t;
}
} LL cal(){
LL sum= ;
LL r = exgcd(A, B, x, y);
LL z = A*B/r; if((+n)%r) return ;
else{
x = x*((+n)/r);
LL d = B/r;
x = (x%d + d) %d;
if(x == )
x += d;
LL remain = n - x*A;
if(remain < ) return ;
else{
sum++;
sum += remain/z;
}
}
return sum;
}
int main(){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
scanf("%I64d",&C);
for(int i = ;i< C;i++){
scanf("%I64d%I64d%I64d",&n, &A, &B);
cout<<cal()<<endl;
}
return ;
}