给定一个以字符串表示的非负整数 num，移除这个数中的 k 位数字，使得剩下的数字最小。注意:num 的长度小于 10002 且 ≥ k。
num 不会包含任何前导零。

import sys # remove k
def removeKdigist(num,k):m = len(num)if m == 0 or m ==k :return '0'if k == 0: return num i = 0while i<m -1 and k>0:print num ,iif num[i] > num[i+1] and num[i+1]!='0' and num[i]!='0':# if num[i] > num[i+1]： 原先得答案 是这样得 这就会导致 0 参与比较，这样得化
#例子中10200 k=2 删除之后 就是200 ，加上 我得改进，是 102 num = num[:i] + num[i+1:]k = k-1 if i >0:i = i-1else:i+=1print num,"aa"print numif k>0:num = num[:m-k]while len(num)>1 and num[0]=='0':num = num[1:]return numstra = "1432219"
k = 3
stra = "10200"
k=1
stra = '10'
k = 2
print removeKdigist(stra,k)

def minNumber(arr):n = len(arr)minNum = -1if len(arr) == 0: return -1left = 0 right = n -1  mid = left#当最左边值 大于 等于 右边的时候 while arr[left] >= arr[right] :# 数组 就剩下两个值if right - left == 1:mid = rightbreakmid =(left + right) /2if arr[left] == arr[mid] and arr[left] == arr[right]:return oderFind(arr,left,right)if arr[mid] >= arr[left]:left = mid elif arr[mid] <=arr[right]:right = mid return arr[mid]
def oderFind(arr,left,right):min = arr[left]while left<=right:if arr[left] < min:min = arr[left]left +=1 return minarr = [1,1,1,1,0,1]
arr = [1,0]
arr = [3,4,5,1,2,3]
arr = [1,1,2,3,4,5,6]
print minNumber(arr) 

import java.util.Scanner;public class foot {public static void main(String[] args) {Scanner sc = new  Scanner(System.in);int le = 1;while(sc.hasNext()) {int n = sc.nextInt();int s = sc.nextInt();int t = sc.nextInt();if(s<le||t<le||s>n||t>n||n<le) System.out.println(-1);else {if(s<t) {if(s==le) {if(n==t) System.out.println(0);else System.out.println(1);}else {if(t-s>1) System.out.println(2);else System.out.println(1);}}else if(s>t) {if(s==n) {if(t==le) System.out.println(0);else System.out.println(1);}else {if(s-t>1) System.out.println(2);else System.out.println(1);}}else {if(le==n) System.out.println(0);else System.out.println(-1);}}}}
}

# -*- coding:utf-8 -*-
#coding:utf-8
import sys
reload(sys)
sys.setdefaultencoding('utf-8')def swap(arr,i ,j):temp = arr[i]arr[i] = arr[j]arr[j] = temp
def array_full(a,left ,right):if left == right-1:print str(a).decode('string_escape')for i in range(left,right):if a[i] in a[left:i]: continue  # 有重复元素得 全排列swap(a,i,left)array_full(a,left+1,right)swap(a,i,left)
stra = ["天","向","上","天"]
array_full(stra,0,len(stra))