Help is needed for Dexter UVA

题目描述：给定一串数字1~n,每次可以选取任意个数字对他们减去任意一个数字x，求将这n个数全部减为0所需的最小处理次数。

思路：我做时算是在边推边找规律吧，推到n为12才突然反应过来：F(n) = F(n / 2) + 1。如1 2 3 4 5 6 7首先将4 5 6 7分别减去4得到1 2 3 0 1 2 3此时相当于将原来的串1 2 3 4 5 6 7 变成了1 2 3，然后将所有的2 3减去2,得到1 0 1 0 1 0 1，最后再将所有的1减去1得到0 0 0 0 0 0 0。原本想用数组保存结果直接输出的，结果发现n最大为1e9，数组开不下，又想了想其实就是分治递归的思想每次给定一个n直接递归求解即可。不过为了节约时间，还是用数组储存了部分结果，其实是没有必要的。直接递归求解时间为30ms,而开数组是340ms，反而更耗时。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int maxn = 2 * 1e8 + 10;const int maxt = 1e6 + 10;const int mod = 10000007;const int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;ll num[maxn];int  n, m, k;ll solve(int n){    if(n == 1) return 1;    return solve(n / 2) + 1;}int main(){    while(scanf("%d", &n) != EOF){        ll ans = solve(n);        printf("%lld\n", ans);    }    return 0;}

下面这段代码，原本想节省时间的，结果反而是多此一举了，更耗时。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int maxn = 2 * 1e8 + 10;const int maxt = 1e6 + 10;const int mod = 10000007;const int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;ll num[maxn];int  n, m, k;void init(){    num[0] = 0;    num[1] = 1;    num[2] = 2;    for(int i = 3; i < maxn; ++i){        num[i] = num[i / 2] + 1;    }}ll solve(int n){    if(n < maxn) return num[n];    return solve(n / 2) + 1;}int main(){    init();    while(scanf("%d", &n) != EOF){        if(n < maxn)printf("%lld\n", num[n]);        else{            ll ans = solve(n);            printf("%lld\n", ans);        }    }    return 0;}