二分搜索 UVALive 6076 Yukari's Birthday (12长春K)

题目传送门

题意：问使得sum (k^i) = n || n -1 (1 <= i <= r) 的min (r*k)组合的r和k 

分析：r的最大不会超过40，枚举r，二分搜索k。注意会爆long long，所以上界需要优化。r = 2开始上界就小于1e6，cyd将后面的范围也求出来了，其实1e6就够用了。

这水题卡了我好久，没有很好分析题目，做不出来就有种无力感，开始烦躁起来。还是题目做得少了，如果这种题做多了，可能看一眼就能做出来了。

/************************************************
* Author        :Running_Time
* Created Time  :2010-1-16 12:18:59
* File Name     :K.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
ll n, r, k;

ll cal(ll x, int y) {
    ll ret = 0, p = x;
    for (int i=1; i<=y; ++i)    {
        ret = ret + p;
        p *= x;
        if (ret > (ll)1e12)   return ret;
    }
    return ret;
}

ll bsearch(ll l, ll r, int m, ll ans)  {
    while (l <= r)   {
        ll mid = (l + r) >> 1;
        ll sum = cal (mid, m);
        if (sum == ans)   return mid;
        else if (sum < ans)   l = mid + 1;
        else    r = mid - 1;
    }
    return 0;
}

int main(void)    {
    while (scanf ("%lld", &n) == 1) {
        r = 1;  k = n - 1;  ll kk;
        for (int i=2; i<=40; ++i)  {
            kk = bsearch (2, 1e6, i, n);
            if (kk == 0)   continue;
            if (1ll * i * kk < r * k) {
                r = i;  k = kk;
            }
        }
        for (int i=2; i<=40; ++i)  {
            kk = bsearch (2, 1e6, i, n - 1);
            if (kk == 0)   continue;
            if (1ll * i * kk < r * k) {
                r = i;  k = kk;
            }
        }
        printf ("%lld %lld\n", r, k);
    }

    return 0;
}