Min Number

总时间限制: 1000ms 内存限制: 32768kB
描述
   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≤|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!






输入
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.
输出
For each test case, output the minimum number we can get after no more than M operations.
样例输入
3 
9012 0
9012 1 
9012 2
样例输出
9012 
1092 

1029

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

int f(char *p_ch , int n) {
	int len = strlen(p_ch);
	int flag = 1;
	for(int i = 0 ; i < n ; i++) {
		int index = 0;
		if(flag) {
			if(i == 0) {
				int tf = 1;
				for(int j = len - 1 ; j > i ; j-- ) {
					if(p_ch[j] - '0' && tf) {
						index = j;
						tf--;
					}
					if(index == 0) continue;
					else {
						if(p_ch[index] > p_ch[j] && p_ch[j] - '0') index = j;
					}
				}//find_min_i(0);
			} else {
				index = len - 1;
				for(int j = len - 1 ; j > i ; j-- ) {
					if(p_ch[index] > p_ch[j] ) index = j;
				}//find_min_i(1);
			}
			if(p_ch[index] < p_ch[i]) {
				int temp;
				temp = p_ch[i];
				p_ch[i] = p_ch[index];
				p_ch[index] = temp;
			}
			else
			{
				if(i == len - 1 )
					break;
				n++;
				continue;
			}
		}
	}
	return 0;
}


int main() {
	char s[1005];
	int size;
	cin >> size;
	while(size--) {
		int n;
		cin >> s >> n;
		f(s,n);
		cout << s << endl;
	}
	//cout << "Hello World!!!" << endl;
	return 0;
}