FZU 1853 Number Deletion

Accept: 17Submit: 61
Time Limit: 1000 mSecMemory Limit : 32768 KB

Problem Description

Given you one n-digital positive integer a,After removing any of them k( k < n) digits, the remaining figures form a new positive integer according to the origin order. For a given n-digital positive integers a and positive integer k. Now ask you to find a algorithm to minimize the remaining integer.

Input

The first line contains one integer t, represents the number of test cases.

Then following t lines,each line contain two numbers a,k (0 < a < 10^1000, k is less than the length of a).

Output

Output the mininum number after the deletion.(the number can not contain leading zero)">it means that we should ignore the leading zeros of the output number

Sample Input

1 178543 4

Sample Output

13

//题目的意思很简单(我英语这么差的人都看得懂,我想应该没人看不懂)

//题目也不难, 注意处理好"0"就OK了,

//但因为多写了一个"=",

//害我折腾了一天..>-.-<

//整得我又是重写, 又手debug查....

//晚上玩了会DNF, 再拿起程序来调试,

//突然...

//发现一组数据有误(终于看到错误的数据了,,,,心里那个激动啊)

//再debug了一遍...

//原来是"<"写成"<="了.....

orz orz orz....

//调试无数次的程序, 看起来有点乱....

#include<iostream> using namespace std; int main() { char c[1002],max; int n,k,i,len,t,zt,zi,tmp,j; bool flag,flag2; scanf("%d",&n); while(n--) { scanf("%s%d",c,&k); len=strlen(c); max=c[0]; zt=zi=0; tmp=k; flag2=flag=false; for(i=0;i<len;i++) { if(c[i]=='0') { if(tmp-i+zt>=0) for(j=zi;j<=i;j++) c[j]='A'; else { zi=i; flag2=true; break; } k=tmp-(i-zt); max=c[i+1]; zt++; zi=i; } else if(max>c[i]) { k--; t=i-1; while(c[t]=='A') t--; c[t]='A'; while(c[t]=='A') t--; if(t<0) max=c[i]; else max=c[t]; i--; } else if(max<c[i]) max=c[i]; if(k==0) break; } if(k!=0) { if(flag2) i=zi-1; else i=len-1; for(;i>=0;i--) { if(c[i]!='A') {c[i]='A';k--;} if(k==0) break; } } for(i=0;i<len;i++) { if(flag) { if(c[i]!='A') printf("%c",c[i]); } else { if(c[i]!='A'&&c[i]!='0') {printf("%c",c[i]);flag=true;} } } if(!flag) printf("0"); printf("/n"); } return 0; }