zoj 3665 数论 二分 两个参数

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4888

两个参数的题，处理方法：枚举小的那个参数，然后二分大的参数

想到二分了，然后两个参数就不会了  然后暴力了下，但是其实K可以很大 所以时间不够

自己写的二分枚举+快速幂程序WA了很久。。。。没明白哪里错了  参考了别人的。。。

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;ll n,ansr,ansk;void test(ll d, ll up, int r){    ll sum;    while(d<up)    {        ll mid=(d+up)/2;        sum=0;        ll know=1;//k^i        int flag=0;        for(int i=1;i<=r;i++)        {            know*=mid;            sum+=know;            if(sum>1000000000000LL || sum<0 )            {                flag=1;                break;            }        }        if(flag)        {            up=mid;            continue;        }        if(sum>n)up=mid;        else            d=mid+1;        if(sum+1 == n || sum==n)        {            if(mid*r < ansr*ansk)                ansr=r,ansk=mid;                break;        }    }}int main(){    IN("zoj3665.txt");    while(~scanf("%lld",&n))    {        ansr=1;        ansk=n-1;//        for(int i=2;i<=64;i++)            test(1ll,1000000,i);        printf("%lld %lld\n",ansr,ansk);    }    return 0;}