HDU3183A Magic Lamp,和NYOJ最大的数一样

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A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2937 Accepted Submission(s): 1145

Problem Description

Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question, and then the genie will realize one of her dreams.
The question is: give you an integer, you are allowed to delete exactly m digits. The left digits will form a new integer. You should make it minimum.
You are not allowed to change the order of the digits. Now can you help Kiki to realize her dream?

Input

There are several test cases.
Each test case will contain an integer you are given (which may at most contains 1000 digits.) and the integer m (if the integer contains n digits, m will not bigger then n). The given integer will not contain leading zero.

Output

For each case, output the minimum result you can get in one line.
If the result contains leading zero, ignore it.

Sample Input

178543 4 1000001 1100001 212345 254321 2

Sample Output

1310123321

我的天，这个题只要细心基本上没什么大问题，哎，，，，做的好辛苦啊，，思路也和NYOJ上贪心专题里的最大的数一样，，，可是把代码复制过来改改交却WA了一遍又一遍，，改的好辛苦，依旧错。。。

看起来都没多大区别：

#include<bits/stdc++.h>using namespace std;const int N=1001;char a[N],b[N];int main(){    int n,i,j,k1,k2,k3,x1,x2;    while(~scanf("%s%d",a,&n))    {        memset(b,'0',sizeof(b));        x1=strlen(a);        if(x1==n)        {            printf("0\n");            continue;        }        else        {            j=k1=k3=0,k2=n,x2=x1-n;            char c;            while(x2--)            {                c=a[k1],k3=k1;                for(i=k1; i<=k2; i++)                {                    if(k1==k2)                    {                        k3=k2;                        break;                    }                    if(a[i]<c)                        c=a[i],k3=i;                  }                b[j++]=c,k1=k3+1,k2++;            }            for(i=0; i<j; i++)                if((b[i]!='0')||(i==j-1))                {                    x2=i;                    break;                }            for(i=x2; i<j; i++)                printf("%c",b[i]);            printf("\n");        }    }    return 0;}

怎么说嘞，只能说做题沉心静气，细心~~

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