暴力枚举

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int n, x, y, i, j;
    int flag, num[11], k=0;

    while(~scanf("%d", &n) && n)
//    while(scanf("%d", &n) != EOF && n)
    {
        if (k) printf("\n"); k++;
        flag = 1;
        for(i=1234; i<50000; i++)
        {
            memset(num,0,sizeof(num));
            x = i*n;
            y = i;
            while(x||y)
            {
                if(!num[y % 10])
                {
                    num[y % 10] = 1;
                    y /= 10;
                }
                else break;
                if(!num[x % 10])
                {
                    num[x % 10] = 1;
                    x /= 10;
                }
                else break;
            }

            for(j=0; j<10; j++)
                if(!num[j])
                    break;
            if(j == 10 && i*n <= 98765)
            {
                printf("%05d / %05d = %d\n", i*n, i, n);
                flag = 0;
            }
        }

        if(flag)
            printf("There are no solutions for %d.\n", n);
    }
    return 0;
}

思路：利用数组下标来存取余数，若abcde与efghi有相同的数字则 数组的和不为10

#include <iostream>
#include <cstdio>

using namespace std;

struct number
{
    int xx;
    int yy;
} num[10010];

int main()
{
    unsigned int k;
    int x, y, m, n;

    while(scanf("%d",&k) != EOF)
    {
        m = 0;
        n = 0;
        for(int y=k+1; y<=2*k; y++)
        {
                x = (int)(k*y/(y-k));
                if(x>=y)
                {
                    if((x*y == (k*y + k*x)))
                    {
                        num[n++].xx = x;
                        num[m++].yy = y;
                    }
                }
        }
        cout << n <<endl;
        for(int i=0; i<n; i++)
            cout << "1/" << k <<" = " <<  "1/" << num[i].xx << " + "<< "1/" << num[i].yy << endl;
    }
    return 0;
}

思路：根据X >= Y ，得到 k < Y <= 2*k 以此来确定Y的枚举范围

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main()
{
    int n, s[20], m = 1;
    long long maxnum;

    while(scanf("%d", &n) != EOF)
    {
        maxnum = 0;
        memset(s,0,sizeof(s));
        for(int i=0; i<n; i++)    //存储数组
            cin >> s[i];

        for ( int i = 0; i < n; ++i )        //枚举子数组的起点和终点
        {
            for ( int j = i; j < n ; ++j )
            {
                long long sum = 1;
                for ( int k = i; k <= j; ++k )
                {
                    sum *= s[k];
                }
                if(sum > maxnum)
                    maxnum = sum;
            }
        }
        printf("Case #%d: The maximum product is %lld.\n\n",m++,maxnum);
    }
    return 0;
}

思路：枚举子数列的起点和终点