hdu 4937 base进制只含3456的base数

http://acm.hdu.edu.cn/showproblem.php?pid=4937

给定一个数n，若这个数在base进制下全由3，4，5，6组成的话，则称base为n的幸运进制，给定n，求有多少个幸运进制。无穷多个的话输出-1，单个位置上超过9用相应的字符表示。

特判n为3~6才会无穷多解

暴力+二分

先特别求出只有两位和用二分求出只有三位表示的对应base数，然后从base = 4开始暴力遍历即可

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include<set>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d:%d",&x,&y)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;

int main() {
    int _;RD(_);LL n;
    for(int tt = 1;tt <= _;++tt){
        int i,j,k;
        scanf("%I64d",&n);
        LL t = n;
        bool flag = 1;
        if(t>=3&&t<=6){
            printf("Case #%d: -1\n",tt);
            continue;
        }
        if (n <= 10){
            printf("Case #%d: 0\n",tt);
            continue;
        }
        LL ans = 0;
        for(i=3;i<=6;++i){
            t = n-i;
            for(j=3;j<=6;++j)
                if(t % j == 0 && t/j > i && t/j > j) ++ans;
        }
        for (i=3;i<=6;++i)
            for (j=3;j<=6;++j)
                for (k=3;k<=6;++k){
                    LL l = 0 , r = (LL)sqrt(n)+1 , mid;
                    while (l < r){
                        mid = (l+r)>>1;
                        t = i + mid*mid*k + mid*j;
                        if (t >= n) r = mid;
                        else l = mid+1;
                    }
                    t = i + l*l*k + l*j;
                    if (t == n && i < l && j < l && k < l) ++ans;
                }
        for (j=4;j;++j){
            t = n;
            bool flag = 1;
            k = 0;
            while (t){
                int tmp = t % j;
                k++;
                if(tmp <3 || tmp > 6){
                    flag = 0;
                }
                t /= j;
            }
            if(k < 4)break;
            if (flag) ++ans;
        }
        printf("Case #%d: %I64d\n",tt,ans);
    }
    return 0;
}