Codeforces Round #402 (Div. 2) A B C sort D二分 (水)

Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10 k .

In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10 k . For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.

Write a program that prints the minimum number of digits to be deleted from the given integer number n , so that the result is divisible by 10 k . The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).

It is guaranteed that the answer exists.

The only line of the input contains two integer numbers n and k (0 ≤ n ≤ 2 000 000 000, 1 ≤ k ≤ 9).

It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.

Print w — the required minimal number of digits to erase. After removing the appropriate w digits from the number n , the result should have a value that is divisible by 10 k . The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).

In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.

2 seconds

256 megabytes

standard input

standard output

 #include <iostream>

 #include <cstdio>

 #include <cstdlib>

 #include <cstring>

 #include <algorithm>

 #include <stack>

 #include <queue>

 #include <cmath>

 #include <map>

 #define ll  __int64

 #define mod 1000000007

 #define dazhi 2147483647

 using namespace  std;

 int n,k;

 struct  node

 {

     int a,b;

     int cha;

 }N[];

 bool cmp(struct node aa,struct node bb)

 {

     return aa.cha<bb.cha;

 }

 int main()

 {

     int ans=;

     scanf("%d %d",&n,&k);

     for(int i=;i<=n;i++)

     {

         scanf("%d",&N[i].a);

     }

      for(int i=;i<=n;i++)

     {

         scanf("%d",&N[i].b);

         N[i].cha=N[i].a-N[i].b;

     }

     sort(N+,N++n,cmp);

     for(int i=;i<=k;i++)

     {

       ans+=N[i].a;

     }

     for(int i=k+;i<=n;i++)

     {

         if(N[i].cha<)

             ans+=N[i].a;

         else

             ans+=N[i].b;

     }

     printf("%d\n",ans);

     return ;

 }

Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.

Sergey gives Nastya the word t and wants to get the word p out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word t : a 1... a | t |. We denote the length of word x as | x |. Note that after removing one letter, the indices of other letters don't change. For example, if t = "nastya" and a = [4, 1, 5, 3, 2, 6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya".

Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word p . Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.

It is guaranteed that the word p can be obtained by removing the letters from word t .

The first and second lines of the input contain the words t and p , respectively. Words are composed of lowercase letters of the Latin alphabet (1 ≤ | p | < | t | ≤ 200 000). It is guaranteed that the word p can be obtained by removing the letters from word t .

Next line contains a permutation a 1, a 2, ..., a | t | of letter indices that specifies the order in which Nastya removes letters of t (1 ≤ a i ≤ | t |, all a i are distinct).

Print a single integer number, the maximum number of letters that Nastya can remove.

In the first sample test sequence of removing made by Nastya looks like this:

"ababcba" "ababcba" "ababcba" "ababcba"

Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".

So, Nastya will remove only three letters.

题意：给你初始字符串 目标字符串 以及对初始字符串执行删除操作的顺序

若对某个字符删除之后无法到达目标状态  则无法删除 并且结束

问最多能删除多少个字符

题解：二分答案  fun()

 #include <iostream>

 #include <cstdio>

 #include <cstdlib>

 #include <cstring>

 #include <algorithm>

 #include <stack>

 #include <queue>

 #include <cmath>

 #include <map>

 #define ll  __int64

 #define mod 1000000007

 #define dazhi 2147483647

 using namespace  std;

 char a[];

 char b[];

 int c[];

 map<int,int> mp;

 int len1,len2;

 bool fun(int x)

 {

     mp.clear();

     for(int i=;i<=x;i++)

     {

         mp[c[i]-]=;

     }

     int exm=;

     int jishu=;

     for(int j=;j<len2;j++)

     {

         while(exm<len1)

       {

          if(a[exm]==b[j]&&mp[exm]==){

             jishu++;

             exm++;

             break;

          }

          else

            exm++;

       }

     }

     if(jishu==len2)

         return true;

     else

         return false;

 }

 int main()

 {

     scanf("%s",a);

     scanf("%s",b);

     len1=strlen(a);

     len2=strlen(b);

     for(int i=;i<len1;i++)

          scanf("%d",&c[i]);

     int l=,r=len1-,mid;

     while(l<=r)

     {

         mid=(l+r)/;

         if(fun(mid))

         {

            l=mid+;

         }

         else

         {

             r=mid-;

         }

     }

     printf("%d\n",r+);

     return ;

 }