hrbust 1387 Shuffle the Digits【贪心】

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Shuffle the DigitsTime Limit: 1000 MSMemory Limit: 65536 KTotal Submit: 165(68 users)Total Accepted: 75(66 users)Rating: Special Judge: NoDescription

Leyni gets an integer n.

He wants to shuffle the digits in the integer n to produce a smallest possible integer without leading zeros. (None of the digits in the integer can be removed or replaced.)

Please help him solve this problem!

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. Then T test cases follow.

For each test case:

Line 1. This line contains an integer n (0 ≤ n ≤ 10⁹).

Output

For each test case:

Line 1. Output the new integer that Leyni wants.

Sample Input

4321

Sample Output

1234

Source哈理工2012春季校赛 - 现场赛Author齐达拉图@HRBUST

思路：

对于没有0出现的数字重组满足这样的条件就能使得输出的数字最小：从出现的最小的数开始输出，对应输出出现的次数，直到所有数字输出完毕。

对于有0出现的数字重组满足这样的条件就能使得输出的数字最小：先输出一个非0的最小的数字，然后输出0，剩下的部分按照没有0的规则输出，模拟代码直接AC:

#include<stdio.h>#include<string.h>using namespace std;int vis[10];int main(){    int t;    scanf("%d",&t);    while(t--)    {        char a[200];        memset(vis,0,sizeof(vis));        scanf("%s",a);        for(int i=0;i<strlen(a);i++)        {            vis[a[i]-'0']++;        }        if(vis[0]==0)        {            for(int i=1;i<10;i++)            {                for(int j=1;j<=vis[i];j++)                printf("%d",i);            }            printf("\n");        }        else        {            int f=0;            for(int i=1;i<10;i++)            {                if(vis[i]!=0)                {                    printf("%d",i);                    vis[i]--;                    break;                }            }            for(int i=1;i<=vis[0];i++)            {                printf("0");            }            for(int i=1;i<10;i++)            {                for(int j=1;j<=vis[i];j++)                printf("%d",i);            }            printf("\n");        }    }}

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