FZU 1853 Number Deletion(贪心)

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Problem 1853 Number Deletion

Accept: 83 Submit: 258
Time Limit: 1000 mSec Memory 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

1178543 4

Sample Output

题意：

删除k个数，使得这个数最小。

思路：

这题简直坑货，让高数位尽量小，于是就要从前向后扫，每次就删当前数前面，比自己大的数；然而注意前导0不输出。

代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;const int N= 1005;char s[N];int vis[N];int main(){    int t,k;    scanf("%d",&t);    while(t--)    {        memset(vis,0,sizeof vis);        scanf("%s%d",s,&k);        int len=strlen(s);        for(int i=1; i<len; i++)        {            for(int j=i-1; j>=0; j--)//从最近的删起            {                if(!vis[j]&&s[j]>s[i]) vis[j]=1,k--;                if(!k) break;            }            if(!k) break;//两次break;        }        if(k)        {            for(int i=len-1; i>=0; i--)            {                if(!vis[i]) vis[i]=1,k--;                if(!k) break;            }        }        int f=1;        for(int i=0; i<len; i++)            if(!vis[i])            {                if(f&&s[i]=='0') continue;                if(s[i]!='0') f=0;                putchar(s[i]);            }            if(f) printf("0");//都是0，则输出0        puts("");    }    return 0;}

另外一种用单调栈做的。代码更简单。时间复杂度O(n)。

#include<cstdio>const int maxn=1010;char num[maxn],q[maxn];int main(){    int t,i,k,tail;    scanf("%d",&t);    while(t--)    {        scanf("%s%d",num,&k);        tail=0;        for(i=0;num[i];i++)        {            while(tail&&num[i]<q[tail-1]&&k)                tail--,k--;            q[tail++]=num[i];        }        while(k)            k--,tail--;        q[tail]=0;        for(i=0;i<tail;i++)            if(q[i]!='0')                break;        if(i==tail)            printf("0\n");        else            printf("%s\n",q+i);    }    return 0;}

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