出题人题解 | #tokitsukaze and Counting#

原题解链接：https://ac.nowcoder.com/discuss/150249

$a n s = r / x - (I - 1) / x$

//
// Created by calabash_boy on 18-10-18.
//
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;

#ifdef __LOCAL_DEBUG__
# define _debug(fmt, ...) fprintf(stderr, "\033[91m[%s %3d]: " fmt "\n\033[0m", \
  __func__,__LINE__, ##__VA_ARGS__)
#else
# define _debug(...) (void(0))
#endif

#define PB(x) push_back(x)
#define rep(i,l,r) for (int i = l,_ = r;i< _;i++)
#define REP(i,l,r) for (int i=l,_=r;i<=_;i++)
#define leave(x) do {cout<<#x<<endl;fflush(stdout);return 0;}while (0);
#define untie do{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);}while (0)

typedef long long LL;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef long double db;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const int inf = 0x3f3f3f3f;
const ll inf_ll = 0x3f3f3f3f3f3f3f3fLL;


/************* header ******************/
const int maxn = 100+50;


int main(){
    int T;
    for (scanf("%d",&T);T;T--){
        ll l,r,x;
        cin>>l>>r>>x;
        cout<<r/x - l/x + (l %x == 0)<<endl;
    }


    return 0;
}