杭电4148-Length of S(n)

Length of S(n)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 767    Accepted Submission(s): 453

Problem Description

A number sequence is defined as following:
S(1)=1,
S(2)=11,
S(3)=21,
S(4)=1211,
S(5)=111221,
S(6)=312211,
……
Now, we need you to calculate the length of S(n).

 

Input

The input consists of multiple test cases. Each test case contains one integers n.
(1<=n<=30)
n=0 signal the end of input.

 

Output

Length of S(n).

 

Sample Input

250

 

Sample Output

26
一道坑爹题，题目看了半天也没看明白，过后来经学长点播终于搞清楚是神马意思了
/*这题每个S(n)是描述S(n-1)值例如：S(1)=1;S(2)=11;即描述S(1)有1个1=11S(3)=21;即描述S(2)有2个1=21S(4)=1211;即描述S(3)有1个2和1个1=1211................*/#include<iostream>//因为题目给的测试数据不大，所以直接打表#include<cstdio>#include<cstring>#include<string>#include<algorithm>const int MAX=5000;char s[31][MAX];//我测试下,到了S(30)左右会长度差不多快到5000using namespace std;//这里S[n][n]是模拟题目S(n)void counnt(){    s[1][0]='1';    s[2][0]='1';    s[2][1]='1';    s[3][0]='2';    s[3][1]='1';    int i,j,k,len;    int num;    for(i=4;i<=30;i++)    {        len=strlen(s[i-1]);        num=1;        k=0;        for(j=1;j<len;j++)        {            if(s[i-1][j]==s[i-1][j-1])//如果前一个数字等于这一个数字那么+1            {                num+=1;            }            else//否则            {                if(num>9)                {                    s[i][k]=num%10+48;                    k+=1;                    num/=10;                    s[i][k]=num+48;                    k+=1;                }                else                {                    s[i][k]=num+48;                    k+=1;                    s[i][k]=s[i-1][j-1];                    k+=1;                    num=1;                }            }        }        if(num>9)        {            s[i][k]=num%10+48;            k+=1;            num/=10;            s[i][k]=num+48;            k+=1;        }        else        {            s[i][k]=num+48;            k+=1;            s[i][k]=s[i-1][j-1];            k+=1;        }    }}int main(){    int n,m;    counnt();    while(cin>>n,n)    {        m=strlen(s[n]);        cout<<m<<endl;    }    return 0;}