Now you are given one non-negative integer n in 10-base notation, it will only contain digits (‘0‘-‘9‘). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].
For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.
Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?
Please note that in this problem, leading zero is not allowed!
The first line of the input contains an integer T (T≤100), indicating the number of test cases.
Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.
每次操作可以换两个数位上的数
最多m次操作
问操作完可以得到的最小的数是多少
贪心,第一位不能是零,然后找最小的数 不断放在最前面的位数。
9012 首位不能是0 所以把其余最小的1 放在首位上 得 1092 ;第二位已经最小 所以把第三位9换成剩下的最小的2 得1029;
原文:http://blog.csdn.net/u013532224/article/details/41753301