Codeforces 165B Burning Midnight Oil 【二分】

题目链接：Codeforces 165B Burning Midnight Oil

B. Burning Midnight Oil
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of n lines of code. Vasya is already exhausted, so he works like that: first he writes v lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , …

The expression is regarded as the integral part from dividing number a by number b.

The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished.

Vasya is wondering, what minimum allowable value v can take to let him write not less than n lines of code before he falls asleep.

Input
The input consists of two integers n and k, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1 ≤ n ≤ 109, 2 ≤ k ≤ 10.

Output
Print the only integer — the minimum value of v that lets Vasya write the program in one night.

Examples
input
7 2
output
4
input
59 9
output
54
Note
In the first sample the answer is v = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task.

In the second sample the answer is v = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that’s even more than n = 59.

题意：有n的作业量，现在一个人第1次完成v的作业量，第二次完成v / k的作业量，第三次完成v / (k*k)的作业量，直到作业量为0停止。问你最小的v。

就是一个二分，脑残还WA一次。

AC代码：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <queue>#define CLR(a, b) memset(a, (b), sizeof(a))#define fi first#define se secondusing namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 2*1e5 +10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;void add(LL &x, LL y) { x += y; x %= MOD; }bool judge(LL o, LL k, LL n) {    LL ans = o;    LL b = k;    if(ans >= n) return true;    while(o >= b) {        ans += o / b;        b = b * k;    }    return ans >= n;}int main(){    LL n, k;    while(scanf("%lld%lld", &n, &k) != EOF) {        LL l = 0, r = n, ans;        while(r >= l) {            LL mid = (l + r) >> 1;            if(judge(mid, k, n)) {                ans = mid;                r = mid - 1;            }            else {                l = mid + 1;            }        }        printf("%lld\n", ans);    }    return 0;}