UVa 1555 - Garland (二分)
题意
求最后一个灯的最低位置。
思路
一开始我还妄想通过推出通项公式来获得答案，推了好久推不出来。看了帆神的题解才知道要二分第二个灯的位置TAT
可以得出通项公式为$C[n] = 2C[n - 1] + 2 - C[n - 2]$
二分第二个灯的位置，并检查是否每项都>=0，如果是的话，就降低位置，继续二分。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-6;
const int MAXN = 1500 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
double C[MAXN], ans;
int n;
bool Check(double mid)
{
C[2] = mid;
for (int i = 3; i <= n; i++)
{
C[i] = 2 * C[i - 1] + 2 - C[i - 2];
if (C[i] < 0) return false;
}
ans = C[n];
return true;
}
int main()
{
int i, j;
double iniHigh;
while (~scanf("%d%lf", &n, &iniHigh))
{
C[1] = iniHigh;
double l = -1, r = MAXN, mid;
while (fabs(r - l) > eps)
{
mid = (l + r) / 2;
if (Check(mid)) r = mid;
else l = mid;
}
printf("%.2f\n", ans);
}
return 0;
}