FOJ 1643 Oaiei's cube II

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 15 Sample Output 1110 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; }