题面

题意

给定三个数a,b,q，再给q组整数l和r，求从l到r有多少个x满足(x%a)%b!=(x%b)%a

思路

分析了一波数据，发现这是个规律题，lcm(a,b)为一个循环节，那就可以直接存一个循环，用后缀存，并且存下一个循环里面满足条件的个数，然后求出l-1和r对应的后缀取差值即可

代码

#include <bits/stdc++.h>
using namespace std;
long long ans[50000];
long long gcd(long long a, long long b)
{
if (a < b)
swap(a, b);
while (a % b != 0)
{
long long t = b;
b = a % b;
a = t;
}
return b;
}
long long gbs(long long a, long long b)
{
return a * b / gcd(a, b);
}
int main()
{
long long t;
cin >> t;
while (t--)
{
long long a, b, q, num = 0;
cin >> a >> b >> q;
int len = gbs(a, b);
for (long long i = 1; i <= len; i++)
{
if (i % a % b == i % b % a)
ans[i] = 0;
else
{
num++;
ans[i] = 1;
}
}
for (long long ss = 0; ss < q; ss++)
{
long long l, r, l1 = 0, r1 = 0;
cin >> l >> r;
l--;
l1 = l / len * num;
for (long long i = 1; i <= l % len; i++)
l1 += ans[i];
r1 = r / len * num;
for (long long i = 1; i <= r % len; i++)
r1 += ans[i];
cout << r1 - l1;
if (ss == q - 1)
cout << endl;
else
cout << ' ';
}
}
return 0;
}