UVa 1554 - Binary Search (枚举)
题意
给一个最后搜索到的数字和搜索的次数,问有哪些区间内的数可以得到这个结果。
思路
一开始还想用BFS推出来,推了好久。后来一想我真是2数据才1W按题目给的二分方法一遍扫描过去O(nlogn)妥妥的。
代码
#include <cstdio>
#include <stack>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#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 X first
#define Y 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;
const double eps = 1e-8;
const int MAXN = 1e4 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int cases = 0;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int A[MAXN];
int N, vis[MAXN], limit;
bool BinarySearch(int x)
{
int p, q, i, L;
p = 0; /* Left border of the search */
q = N-1; /* Right border of the search */
L = 0; /* Comparison counter */
while (p <= q) {
i = (p + q) / 2;
++L;
if (A[i] == x) return L == limit;
if (x < A[i])
q = i - 1;
else
p = i + 1;
}
return false;
}
int main()
{
//ROP;
int i, j, target;
for (i = 0; i <= MAXN; i++) A[i] = i;
while (~scanf("%d%d", ⌖, &limit))
{
bool flag = false;
MS(vis, 0);
for (N = 1; N <= 10000; N++)
if (BinarySearch(target)) vis[N] = 1;
int pre;
vector<pii> ans;
for (i = 1; i <= N; i++)
{
if (vis[i] && !vis[i - 1]) pre = i;
if (vis[i] && !vis[i + 1]) ans.PB({pre, i});
}
printf("%d\n", SZ(ans));
for (i = 0; i < SZ(ans); i++) printf("%d %d\n", ans[i].X, ans[i].Y);
}
return 0;
}