HDOJ 1058

动态规划题，所有的humble number表示为 n = (2^i)*(3^j)*(5^k)*(7^l)。

需要注意的是，i,j,k,l也只能是humble number，因此只能从已产生的humble number数组中选取。

#include <cstdio>
long long int a[5843];
long long int t1,t2,t3,t4;

long long int minimum(int a1, int a2, int a3, int a4)
{
    long long int ret = a1;
    if (a2 <= ret) {
        ret = a2;
    }
    if (a3 <= ret) {
        ret = a3;
    }
    if (a4 <= ret) {
        ret = a4;
    }
    if (ret == a1)
        t1++;
    if (ret == a2)
        t2++;
    if (ret == a3)
        t3++;
    if (ret == a4)
        t4++;
    return ret;
}
void init()
{
    a[1] = 1;
    t1 = t2 = t3 = t4 = 1;
    for(int i=2; i<=5842; ++i) {
        a[i] = minimum(a[t1]*2, a[t2]*3, a[t3]*5, a[t4]*7);
    }
}
int main()
{
#ifdef LOCAL
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
#endif
    int n;
    
    init();

    while ( (scanf("%d",&n) == 1) && (n != 0)) {
        if (n%100 == 11 || n%100 == 12 || n%100 == 13) {
            printf("The %dth humble number is %d.\n",n,a[n]);
            continue;
        }
        switch(n%10) {
            case 1:
                printf("The %dst humble number is %d.\n",n,a[n]);
                break;
            case 2:
                printf("The %dnd humble number is %d.\n",n,a[n]);
                break;
            case 3:
                printf("The %drd humble number is %d.\n",n,a[n]);
                break;
            default:
                printf("The %dth humble number is %d.\n",n,a[n]);
                break;
        }
    }
    return 0;
}