POJ 1426 Find The Multiple

解题思路:

1）每5位为一个基数，枚举5位所有只包含0,1的数ans[]，map[i][j]保存(ans[j]*100000^i)%n的值

2）BFS计算所有组合出现的模，直到出现0

#include <iostream>
using namespace std;

#define MAXN 201
#define radix 100000
const int eu[] = {1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,10000,10001,10010,10011,10100,10101,10110,10111,11000,11001,11010,11011,11100,11101,11110,11111};
int main()
{
    bool visit[MAXN];
    int id[MAXN][2],mod[20][32], q[MAXN], next[MAXN], ans[20];
    int n, p, i, j, k, t, iter, e;
    while (scanf("%d", &n)&&n)
    {
        for(i=0;i<MAXN;i++)visit[i]=0,next[i]=-1;
        memset(ans, 0, sizeof(ans));
        for(i = 0;i < 20; i++)
        {
            for (j=0;j<31;j++)
            {
                if(i>0)mod[i][j] = (mod[i-1][j]*radix)%n;
                else mod[i][j] = eu[j] % n;
                if(i==0&&!visit[mod[i][j]])visit[mod[i][j]]=true,id[mod[i][j]][0]=i,id[mod[i][j]][1]=j;
            }
        }
        for(p=i=0;i<n;i++)if(visit[i])q[p++]=i;
        for (i=1;i<20&&!visit[0];i++)
        {
            for(e=p,j=0;j<31&&!visit[0];j++)
            {
                t = mod[i][j];
                if(t==0)
                    int b = 0;
                if(!visit[t])visit[t]=true,id[t][0]=i,id[t][1]=j,q[p++]=t;
                for(k=0;k<e&&!visit[0];k++)
                {
                    t = (mod[i][j] + q[k]) % n;
                    if(t==0)
                        int b = 0;
                    if(!visit[t])visit[t]=true,id[t][0]=i,id[t][1]=j,next[t]=q[k],q[p++]=t;
                }
            }
        }
        for (iter=id[0][0],i=0;i!=-1;i=next[i])
        {
            j=id[i][0],k=id[i][1];
            p=iter-j;
            ans[p]=eu[k];
        }
        printf("%d",ans[0]);
        for (i=1;i<=iter;i++)
            printf("%05d", ans[i]);
        printf("\n");
    }
    return 0;
}