hdu 4937 base进制只含3456的base数

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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;}

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