EOJ Monthly 2019.2 (based on February Selection) D.进制转换

题目链接：

　　https://acm.ecnu.edu.cn/contest/140/problem/D/

题目：

思路：

　　我们知道一个数在某一个进制k下末尾零的个数x就是这个数整除k^x,这题要求刚好末尾有m个0，还需要除去高位为0的情况，因此这题答案就是r / k^x-(l-1)/k^x-(r/k^x+1-(l-1)/k^x+1)。

代码实现如下：

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 2e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int t;
LL l, r, k, m;

int main(){
    scanf("%d", &t);
    while(t--) {
        scanf("%lld%lld%lld%lld", &l, &r, &k, &m);
        int flag = 1;
        LL ans = 0, tmp = 1;
        for(int i = 1; i <= m; i++) {
            if(r / tmp < k) {
                flag = 0;
                break;
            }
            tmp = tmp * k;
        }
        if(!flag) {
            printf("0
");
            continue;
        }
        ans = r / tmp - (l - 1) / tmp;
        if(r / tmp >= k) {
            tmp = tmp * k;
            ans -= r / tmp - (l - 1) / tmp;
        }
        printf("%lld
", ans);
    }
    return 0;
}