FZU

 Problem 2111 Min Number

Accept: 980    Submit: 1923
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

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!

 Input

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.

 Output

For each test case, output the minimum number we can get after no more than M operations.

 Sample Input

39012 09012 19012 2

 Sample Output

901210921029

 Source

“高教社杯”第三届福建省大学生程序设计竞赛

有次数限制的选择排序

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int T,m;char num[1010];int main(){    scanf("%d",&T);    while(T--){        scanf("%s%d",num,&m);        int len = (int)strlen(num),sum = 0;        for(int i = 0; i < len; i++) sum += num[i] - '0';        if(m*sum==0) {puts(num);continue;}        int mi = '9' + 1,pos = -1;        for(int i = 1; i < len; i++){            if(num[i] == '0') continue;            if(mi >= num[i]) mi = num[i],pos = i;        }        if(num[0]=='0'||num[0]>mi) {swap(num[0],num[pos]);m--;}        for(int i = 1; i < len && m; i++){            mi = i;            for(int j = i + 1; j < len; j++){                if(num[mi] >= num[j]) mi = j;            }            if(num[mi]!=num[i]) {swap(num[i],num[mi]);m--;}                    }        puts(num);    }    return 0;}