Y-25

Description

The look and say sequence is defined as follows. Start with any string of digits as the first element in the sequence. Each subsequent element is defined from the previous one by "verbally" describing the previous element. For example, the string 122344111 can be described as "one 1, two 2's, one 3, two 4's, three 1's". Therefore, the element that comes after 122344111 in the sequence is 1122132431. Similarly, the string 101 comes after 1111111111. Notice that it is generally not possible to uniquely identify the previous element of a particular element. For example, a string of 112213243 1's also yields 1122132431 as the next element.

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Input

The input consists of a number of cases. The first line gives the number of cases to follow. Each case consists of a line of up to 1000 digits.

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Output

For each test case, print the string that follows the given string.

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Sample Input

3
122344111
1111111111
12345

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Sample Output

1122132431
101
1112131415

题意描述：

给你一串数，让你以口语的形式来表达，例如：1111111111即10个1那么就输出 101.

解题思路：

刚开始想想的太复杂，即想运用预处理的方法，发现没有必要。主要困难在输出前面有过的数字的个数，所以不能用预处理的方法。

解题细节：

strlen函数需要调用头文件string.h，为方便起见，可以使用#include<bits/stdc++.h>,其中包含

#include <iostream> #include <cstdio> #include <fstream> #include <algorithm> #include <cmath> #include <deque> #include <vector> #include <queue> #include <string> #include <cstring> #include <map> #include <stack> #include <set>.......

代码：

#include<bits/stdc++.h>using namespace std;int main(){    int n;    char s[1001];    int i,j;    int sum;    cin>>n;    while(n--)    {        cin>>s;        sum=1;        j=strlen(s);        for(i=1;i<=j;i++)        {            if(s[i-1]==s[i])            {                sum++;            }            else            {                cout<<sum<<s[i-1];                sum=1;            }        }        cout<<endl;    }    return 0;

心得：

仔细思考，不要将问题想的复杂化，善于思考。