pat 1140
1140 Look-and-say Sequence (20分)
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
题意:给定一个数,按题目要求生成后n个数,输出第n个数
思路:简单字符串处理
代码如下:(注释部分是使用字符串数组来处理的,最后一个测试点会超时或段错误,欢迎评论区探讨)
#include<cstdio> #include<iostream> #include<string> using namespace std; int main(){ string s; string temp=""; int n; cin>>s>>n; for(int i=1;i<n;i++){ temp=""; for(int j=0;j<s.length();j++){ int count=1; int k=j; while(s[j+1]==s[k]&&j+1<s.length()){ j++; count++; } temp+=s[k]+to_string(count); } s=temp; } cout<<s; return 0; } //#include<cstdio> //#include<string> //#include<iostream> //#include<cstring> //using namespace std; //char str[5000]; //char temp[5000]; //int main(){ // int n; // scanf("%s %d",str,&n); // if(n==1){ // printf("%s",str); // return 0; // } // for(int i=0;i<n-1;i++){ // int pos=0; // for(int j=0;j<strlen(str);j++){ // int k=j; // int count=1; // while(str[j+1]==str[k]&&j+1<strlen(str)){ // j++; // count++; // } // temp[pos++]=str[k]; // temp[pos++]=count+'0'; // } // temp[pos]='\0'; // memcpy(str,temp,strlen(temp)); //// for(int j=0;j<=strlen(temp);j++){ //// str[j]=temp[j]; //// } //// // } // printf("%s\n",str); // return 0; //}