Lucky Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1936 Accepted Submission(s): 590
Problem Description
“Ladies and Gentlemen, It’s show time! ”
“A thief is a creative artist who takes his prey in style... But a detective is nothing more than a critic, who follows our footsteps...”
Love_Kid is crazy about Kaito Kid , he think 3(because 3 is the sum of 1 and 2), 4, 5, 6 are his lucky numbers and all others are not.
Now he finds out a way that he can represent a number through decimal representation in another numeral system to get a number only contain 3, 4, 5, 6.
For example, given a number 19, you can represent it as 34 with base 5, so we can call 5 is a lucky base for number 19.
Now he will give you a long number n(1<=n<=1e12), please help him to find out how many lucky bases for that number.
If there are infinite such base, just print out -1.
Input
There are multiply test cases.
The first line contains an integer T(T<=200), indicates the number of cases.
For every test case, there is a number n indicates the number.
Output
For each test case, output “Case #k: ”first, k is the case number, from 1 to T , then, output a line with one integer, the answer to the query.
Sample Input
2 10 19
Sample Output
Case #1: 0 Case #2: 1
Hint
10 shown in hexadecimal number system is another letter different from ‘0’-‘9’, we can represent it as ‘A’, and you can extend to other cases.
Author
UESTC
Source
2014 Multi-University Training Contest 7
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题意：
给你一个十进制的数字n，然后问你转化为某一进制后它的每一位的数字只可能为3，4，5，6.求这种符合条件的进制有多少种。无穷多输出-1.
分析：
我们枚举n转化为某进制时候有几位数。
对于一位的状况，3,4,5,6明显是-1
对于二位，解方程a*x+b=n
对于三位，解方程a*x^2+b*x+c=n
其他的从4进制开始枚举即可
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define LL long long
using namespace std;
int kcase = 1;
bool judge(LL base, LL n)
{
while(n)
{
if(n % base < 3 || n % base > 6)
return false;
n /= base;
}
return true;
}
void solve()
{
LL n;
scanf("%lld", &n);
printf("Case #%d: ", kcase++);
if(n >= 3 && n <= 6)
{
printf("-1\n");
return ;
}
LL ans = 0;
for(LL a = 3; a <= 6; a++)
{
for(LL b = 3; b <= 6; b++)
{
if((n-a) % b) continue;
if((n-a) / b > max(a, b))
ans++;
}
}
for(LL c = 3; c <= 6; c++)
{
for(LL b = 3; b <= 6; b++)
{
for(LL a = 3; a <= 6; a++)
{
LL C = c - n;
double temp = sqrt(b*b - 4*a*C);
if(temp == (LL)temp)
{
LL d = (LL)temp;
if((d-b) % (2*a) == 0)
{
LL x = (d-b) / (2*a);
if(x > max(max(a, b), c))
ans++;
}
}
}
}
}
for(LL base = 2; base * base * base <= n; base++)
if(judge(base, n))
ans++;
printf("%lld\n", ans);
}
int main()
{
int t;
scanf("%d", &t);
while(t--){
solve();
}
return 0;
}
这个代码一直WA，不知道为什么
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int N=1e5+5;
ll n,m;
int main() {
int t;
scanf("%d",&t);
int kase=0;
while(t--)
{
scanf("%lld",&n);
if(n>=3&&n<=7)
{
printf("Case #%d: -1\n",++kase);
continue;
}
ll ans=0;
///2位数
for(ll i=3;i<=6;i++)
for(ll j=3;j<=6;j++)
{
if((n-j)%i==0)
{
ll temp=(n-j)/i;
if(temp>max(i,j))
ans++;
}
}
///三位数
for(ll i=3;i<=6;i++)
for(ll j=3;j<=6;j++)
for(ll k=3;k<=6;k++)
{
ll c=k-n;
double d=sqrt(j*j-4*i*c);
if(d==(ll)d)
{
if(((ll)d-j)%(2*i)==0)
{
ll x=((ll)d-j)/(2*i);
if(x>max(max(i,j),k))
{
ans++;
}
}
}
}
///其他位数
for(ll i=2;i*i*i<=n;i++)///枚举所有进制
{
ll tt=n;
// int flag=1;
while(tt)
{
if(tt%i<3||tt%i>6)
{
//flag=0;
break;
}
tt/=i;
}
if(tt==0)
{
ans++;
}
}
printf("Case #%d: %lld\n",++kase,ans);
}
return 0;
}