PAT A1140. Look-and-say Sequence (20)

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本文介绍了一种基于描述的整数序列——Look-and-say序列，并提供了一个C++实现来计算给定数字D的第N项。该序列通过描述前一项中每个数字出现的次数和顺序生成下一项。

Look-and-say Sequence (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Look-and-say sequence is a sequence of integers as the following:

D, D1, D111, D113, D11231, D112213111, …
where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1’s, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D.

Input Specification:

Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (<=40), separated by a space.

Output Specification:

Print in a line the Nth number in a look-and-say sequence of D.

Sample Input:
1 8
Sample Output:
1123123111

#include<cstdio>
#include<cstring>

int n;
char d;
char s[100][200000];
int main(){
    scanf("%c %d",&d,&n);
    s[0][0]=d;
    for(int i=1;i<=n;i++){
        int j=0;
        int k=0;
        char c=s[i-1][0];
        int cnt=0;
        while(s[i-1][j]!='\0'){
            if(s[i-1][j]==c)cnt++;
            else{
                s[i][k]=c;
                s[i][k+1]=cnt+'0';
                k+=2;
                c=s[i-1][j];
                cnt=1;
            }
            j++;
        }
        s[i][k]=c;
        s[i][k+1]=cnt+'0';
    }
    printf("%s",s[n-1]);
    return 0;
}