C. Yet Another Counting Problem(数论&取模)
思路:
AC代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a,b,q,c;
ll fun(ll x){
ll cnt=x/c;
ll re=x%c+1;
return x-cnt*b-min(re,b);
}
int main(){
int t;
cin>>t;
while(t--){
cin>>a>>b>>q;
if(a>b) swap(a,b);
c=a/__gcd(a,b)*b;
while(q--){
ll l,r;
cin>>l>>r;
printf("%lld\n",fun(r)-fun(l-1));
}
}
return 0;
}