URAL1095[Nikifor 3]

恩，因为一定有1，2，3，4。而且1234可以组合出mod7的值为0到6的数，所以就先留下一组1234，其他直接输出（注意0不能开头），然后用1234来将mod7的值配成0。

CODE:

/*

PROGRAM: $PROGRAM

AUTHOR: Su Jiao

DATE: 2010-4-15

DESCRIPTION:

$DESCRIPTION

*/

#include <stdio.h>

#include <string.h>

 

#define MAXLEN 20

 

int N;

char num[MAXLEN+1];

int counter[10];

 

int main()

{

    scanf(“%d\n”,&N);

    for (int i=0;i<N;i++)

    {

        scanf(“%s\n”,num);

        memset(counter,0,sizeof(counter));

        for (char* j=num;*j;j++)

            counter[*j-‘0’]++;

        int mod7=0;

        for (int j=9;j>4;j–)

            for (int k=0;k<counter[j];k++)

            {

                printf(“%d”,j);

                mod7=(mod7*10+j)%7;

            }

        for (int j=4;j>0;j–)

            for (int k=1;k<counter[j];k++)

            {

                printf(“%d”,j);

                mod7=(mod7*10+j)%7;

            }

        for (int k=-4;k<counter[0];k++)

            mod7=(mod7*10)%7;

        int mod1234=1;

        for (int k=0;k<counter[0];k++)

            mod1234=(mod1234*10)%7;

        bool printed=false;

        for (int i1=1;i1<=4&&!printed;i1++)

            for (int i2=1;i2<=4&&!printed;i2++)

                for (int i3=1;i3<=4&&!printed;i3++)

                    for (int i4=1;i4<=4&&!printed;i4++)

                        if (i1!=i2&&i1!=i3&&i1!=i4

                            &&i2!=i3&&i2!=i4

                            &&i3!=i4)

                           if (((i1*1000+i2*100+i3*10+i4)*mod1234+mod7)%7==0)

                           {

                              printf(“%d%d%d%d”,i1,i2,i3,i4);

                              printed=true;

                           }

        for (int k=0;k<counter[0];k++)

            printf(“0”);

        printf(“\n”);

    }

    return 0;

}