FZU

先上题目：

Problem 2062 Suneast & Yayamao

Accept: 146    Submit: 319
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 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

There are multiple test cases. Each test case has an integer n(1<=n<=2147483647) in a single line.

 Output

For each case, output a single integer in a line indicate the number of pieces of money people should bring with to buy a Yayamao whose price is random from 1 to n.

 Sample Input

1

2

5

 Sample Output

1

2

3

 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：

#include <cstdio>
#include <cstring>
#define LL long long
#define MAX 2147483647
using namespace std;


int main()
{
    int n;
    int tot;
    //freopen("data.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        tot=0;
        while(n){
            tot++;
            n=n>>1;
        }
        printf("%d
",tot);
    }
    return 0;
}