[Swust OJ 403]--set of deletions

Title Link: http://acm.swust.edu.cn/problem/403/

Time limit (ms): Memory Limit (KB): 65535 DescriptionA collection has the following elements: 1 is the collection element, if p is the element of the set, then 2 * p +1,4*p+5 is also the element of the collection, take out the smallest element of k in this set, in order to combine from small to large to a multi-digit, it is required to remove the number m digits, so that the remaining number is the largest, Programmed output before and after deletion of multiple digits.
Note: There are no cases where all the numbers are deleted InputEnter only one row, the value of k,m, and k,m are less than or equal to 30000. OutputThe output is two lines, the first behavior is deleted before the number, and the second behavior is deleted after the number. Sample Input

5 4

Sample Output

13791595

Thinking of solving problems: Deal with bad words on the pit dad, for int conversion string Convenient lazy, Baidu Sstream head file usage, also learned to ~ ~ ~ (1) Use the collection information in the problem to get the smallest number (2) Sstream,int converted to string to be deleted sequence (3) Greedy---get deleted after the maximum value code is as follows (slightly rubbed):

1#include <iostream>2#include <string>3#include <cstdio>4#include <algorithm>5#include <sstream>//int Turn string6 using namespacestd;7typedefLong LongLL;8 intMain () {9     intLen, M, front, CNT, end;TenLL x[60010] = {0,1}, a, B; One      while(Cin >> Len >>m) { A         inti =1, j =1, K; -          for(k =2; K <= Len; k++){ -A = x[i] *2+1; theb = x[j] *4+5; -             if(A >b) { -J + +; -X[K] =b; +             } -             Else if(A = =b) { +i++; AJ + +; atX[K] =A; -             } -             Else{ -i++; -X[K] =A; -             } in         } -         strings ="", ans =""; to          for(i =1; I <= Len; i++){ + StringStream SS; -             stringstr; theSS <<X[i]; *SS >>str; $s + =str;Panax Notoginseng         } -cout << S <<Endl; the         string:: Iterator it =S.begin (); +S.insert (IT,'9');//effective sequence starting from 1 to prevent front from crossing ALen =s.size (); theFront = CNT =0, end =1; +          while(End <= len && cnt! =m) { -             if(S[end] <=S[front]) $S[++front] = s[end++]; $             Else{ -front--; -cnt++; the             } -         }Wuyi          while(End <=len) theS[++front] = s[end++]; -          for(i =1; i < len-m; i++) Wucout <<S[i]; -cout <<Endl; About     } $     return 0; -}

View Code

[Swust OJ 403]--set of deletions