uva_11384_Help is needed for Dexter(貪心)
2013-04-09 20:59
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題意: 給你一個序列S = {1, 2, 3, ..., n}, 每次可以從中選擇一些元素刪除特定的數字,當然所刪除的數字必須小於等於取出來的元素,最後使得S = {0},的最小刪除次數 分析: 不難想出一個DP方法:設狀態f[1,2,3...,n]爲所求 當 n = 1 時,f[1] = 1. 當 n = 2 時,f[1,2] = max(f[1])+1 當 n = 3 時,f[1,2,3] = max(f[1,2], f[1])+1 ... 狀態轉移:f[1,2,3,...,n] = max(f[1,2,3,...,n-1], f[1,2,3,...,n-2], f[1,2,3,...,n-3] ... f[1,2,3,...,n/2])+1 當然拉,n越大所需要的次數越多,再加上這裏n特別大(10^9),DP無論於時間還是空間上都不符合要求, 所以得出f[1,2,3...n] = f[1,2,3,...n/2]+1的貪心算法 Code: #include <set> #include <map> #include <cmath> #include <ctime> #include <stack> #include <queue> #include <deque> #include <vector> #include <cstdio> #include <bitset> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; #define DIR 4 #define DIM 2 #define STATUS 2 #define MAXN 1000 + 10 #define MAXM 100000 + 10 #define oo (~0u)>>1 #define INF 0x3F3F3F3F #define REPI(i, s, e) for(int i = s; i <= e; i ++) #define REPD(i, e, s) for(int i = e; i >= s; i --) static const double EPS = 1e-5; inline int cal(int n) { if( 1 == n ) { return 1; } return cal(n>>0x1)+1; } int main(int argc, char const *argv[]) { #ifndef ONLINE_JUDGE freopen("test.in", "r", stdin); #endif int n; while( ~scanf("%d", &n) ) { printf("%d\n", cal(n)); } return 0; }
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