FOJ 1643 Oaiei's cube II

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Oaiei's cube II

Time Limit:1s	Memory limit:32M
Accepted Submit:16	Total Submit:31

One day, oaiei plays with his cubes again. He writes “0” or “1” in his N cubes. Then he arranges these cubes to form a series of binary number. But, he is bored with this simple game, so he restricts the game as follows:

”0” is the first number in the sequence, and it is index is 1.
there are not more than three “1”s in the number .
all numbers in the sequence is increased and the index of each number is from 1

So the first 16 number is : 0，1，10，11，100，101，110，111，1000，1001，1010，1011，1100，1101，1110，10000.

Now, your task is to find the binary number according the index.

Input

There are multiple test cases. For each test case, there is an integer N, denoting the index (1<=N<=1.5 * 10 ^ 9) .

Output

For each test case, you should output the binary number according the index.

Sample Input

Sample Output

Original: FZU 2008 Summer Training II--Combinatorics

1.先生成小数据.(用暴力)
下面是1~32的数据：
1:0
2:1
3:10
4:11
5:100
6:101
7:110
8:111
9:1000
10:1001
11:1010
12:1011
13:1100
14:1101
15:1110
16:10000
17:10001
18:10010
19:10011
20:10100
21:10101
22:10110
23:11000
24:11001
25:11010
26:11100
27:100000
28:100001
29:100010
30:100011
31:100100
32:100101
通过位数分类,f[i]表示i位所有的种数
f[1]=2;f[2]=2;f[3]=4;f[4]=7;f[5]=11...f[n]=f[n-1]+n-1;
然后把位对应的种数加起来.使得给出一个数字可以迅速在o(n)内发现它的位数bit，这对于解题来说是一

大步
2.根据题目的条件知道表达式中最多只有3个1，那么对于任何一个数，第一个数字是1应该没什么疑问了(

从左边开始数).然后实际上后面的bit-1位中最多只能出现2个1
3.先考虑一个1的情况,很显然是bit-1种
例如
5:100
6:101
7:110
8:111
显然101,110是满足后面只有一个一的条件的,是3-1=2个.
然后很明显如果所给的数确定了位数以后减去sum[bit],如果是介于1..bit-1,那很明显就可以马上得到答

案.
4.考虑2个1的情况，和考虑1的情况类似，具体见代码不累述了.
总结:暴力生成数据观察规律，模拟解答

#include <iostream> 
using namespace std;
int a[2100]={0,2,2},sum[2100]={0,2};
int main()
{
    int i,n,bit,k,bit1;
    for(i=3;i<=2100;i++)
        a[i]=a[i-1]+i-1;
    for(i=2;;i++)
    {sum[i]+=sum[i-1]+a[i];
        if(sum[i]>=1500000000) break;
    }
    while(scanf("%d",&n)!=EOF)
        {
            for(i=1;i<=2088;i++)
                if(n-sum[i]<=0)
                    break;
            bit=i;
            if(bit==1)
                {
                    cout<<n-1<<endl;
                    continue;
                }
            if(n==sum[bit-1]+1)
                {
                    printf("1");
                    for(i=1;i<=bit-1;i++)
                        printf("0");
                    printf("/n");
                    continue;
                }
            if(n==sum[bit-1]+2)
                {
                    printf("1");
                    for(i=1;i<=bit-2;i++)
                        printf("0");
                    printf("1/n");
                    continue;
                }
            printf("1");
            n-=sum[bit-1];
            n-=2;
            for(i=1;;i++)
                if(n>i+1)
                    n-=i+1;
                else
                    {
                        bit1=i+1;
                        break;
                    }
            bit--;
            for(i=1;i<=bit-bit1;i++)
                printf("0");
            printf("1");
            bit=bit1-1;
            if(n<=2)
                {
                    
                    bit--;
                    for(i=1;i<=bit;i++)
                        printf("0");
                    printf("%d",n-1);
                    printf("/n");
                    continue;
                }
            n-=2;
            for(i=1;i<=bit-n-1;i++)
                printf("0");
            printf("1");
            for(i=1;i<=n;i++)
                printf("0");
            putchar('/n');
        }
   return 0;
}