首页 > 其他 > 详细

C. Adding Powers

时间:2020-03-28 00:14:22      阅读:52      评论:0      收藏:0      [点我收藏+]

题目链接:https://codeforces.com/contest/1312/problem/C

 

想法:

读题之后就会发现其实就是把一个数转化成 k 进制并且 k 进制中的某一位只可以使用一次

 

#pragma GCC optimize(3,"Ofast","inline")//O3优化
#pragma GCC optimize(2)//O2优化
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <cstring>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)

const double eps = 1e-10;
const int maxn = 2e6 + 10;
const LL mod = 1e9 + 7;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

int vis[100];
int main() {
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while (T--) {
        memset(vis,0,sizeof(vis));
        int n;
        LL k;
        cin >> n >> k;
        bool fl = false;
        for (int i = 1;i <= n;i++) {
            int cnt = 0;
            LL a;
            cin >> a;
            if (a == 0 || fl)
                continue;
            while (1) {
                if (a % k == 1) {
                    if (!vis[cnt])
                        vis[cnt] = 1;
                    else {
                        fl = true;
                        break;
                    }
                }
                else if (a % k > 1) {
                    fl = true;
                    break;
                }
                a = a / k;
                cnt++;
                if (a == 0)
                    break;
            }
        }
        if (fl)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}

 

C. Adding Powers

原文:https://www.cnblogs.com/-Ackerman/p/12585010.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!