ACM_下一个排列

The Next Permutation

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

For this problem, you will write a program that takes a (possibly long) string of decimal digits, and outputs the permutation of those decimal digits that has the next larger value (as a decimal number) than the input number. For example: 
123 -> 132 
279134399742 -> 279134423799 
It is possible that no permutation of the input digits has a larger value. For example, 987. 
译文：对于这个问题，你将编写一个程序，它接受一个（可能很长的）十进制数字串，并输出具有比输入数字更大的值（十进制数）的那些十进制数字的排列。例如：
123  - > 132 
279134399742  - > 279134423799 
输入数字的排列可能没有更大的值。例如，987。

Input:

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by up to 80 decimal digits which is the input value.
译文：第一行输入包含一个整数P，（1≤P≤1000），这是后面的数据集的数量。每个数据集都是一行，其中包含数据集编号，后跟一个空格，然后是最多80个十进制数字，即输入值。

Output:

For each data set there is one line of output. If there is no larger permutation of the input digits, the output should be the data set number followed by a single space, followed by the string BIGGEST. If there is a solution, the output should be the data set number, a single space and the next larger permutation of the input digits.
译文：对于每个数据集有一行输出。如果输入数字没有大的排列，则输出应该是数据集编号，后跟一个空格，然后是字符串BIGGEST。如果有解决方案，输出应该是数据集编号，一个空格和下一个较大的输入数字排列。

Sample Input:

3 
1 123 
2 279134399742 
3 987

Sample Output:

1 132
2 279134423799
3 BIGGEST
解题思路：80个10进制数字，即80位数，基本数据类型都不能表示，但C++万能的STL中有next_permutation，用它即可解决此题是否有下一个排列。
str.begin()和str.end() 可以快速访问到字符串的首字符和尾字符。如果有下一个排列，next_permutation(str.begin(), str.end())为真值，否则为假值。
AC代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int p,n;string str;
    cin>>p;
    for(int i=1;i<=p;++i){
        cin>>n>>str;
        cout<<n<<' ';
        if(next_permutation(str.begin(),str.end()))cout<<str<<endl;
        else cout<<"BIGGEST"<<endl;
    }
    return 0;
}