FZU - 2062 - Suneast & Yayamao

 Problem Description

Yayamao is so cute that people loves it so much.

Everyone wants to buy Yayamao from Suneast (a business man who sells Yayamao).

 

 

Suneast is a strange business man. He sells Yayamao in a random price from 1, 2, 3, 4, 5…, n.

Suneast is also a lazy business man. He never looks for a change. But people can’t but Yayamao with a lower price, that say people must pay exact money for Yayamao.

Now, we want to know how many pieces of money people should bring with to buy a Yayamao with the exactly price.

 Input

 Output

 Sample Input

 Sample Output

 Hint

In test case 1: people can bring 1 piece of money: 1

In test case 2: people can bring 2 pieces of money: (1, 1) or (1, 2)

In test case 3: people can bring 3 pieces of money: (1, 1, 3) or (1, 2, 2) ….

　　题意：给你一个数n，问你至少用多少个数，只用这些数就可以表示1~n的所有数。

　　比如5：1=1

　　　　　  2=1+1

　　　　　  3=1+2

　　　　　  4=2+2

　　　　     5=1+1+3

　　当然可能还有很多的方法来表示1~5的数，不过一定需要三个。

　　其实我们可以这样分析，将这个数n化成二进制，假如它有l位，那么对于一个l位的二进制数，我们需要用l位二进制才可以将它表示出来，同时比它小的数，我们可以用不超过l位的数来保存表示它们，换言之，我们可以用2^0,2^1，2^2···2^k···这些数来表示从1~n的数，至于k等于多少，那就需要看n的二进制有多少位了。

　　这也是背包问题里面的多重背包的基础。

上代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #define LL long long
 4 #define MAX 2147483647
 5 using namespace std;
 6 
 7 
 8 int main()
 9 {
10     int n;
11     int tot;
12     //freopen("data.txt","r",stdin);
13     while(scanf("%d",&n)!=EOF){
14         tot=0;
15         while(n){
16             tot++;
17             n=n>>1;
18         }
19         printf("%d\n",tot);
20     }
21     return 0;
22 }

2062