ZOJ 1095 Humble Numbers

参考别人的,就是输出第n个丑数。。。

要先打表,丑数:2、3、5、7的乘积

#include <stdio.h>
#define MAX 2000000000
long long a[5842];
int cmp(const void * x, const void *y)
{
    if( *(int*)x > *(int*)y ) return 1;
    else return -1;
}
int main(void)
{
    long long i, j, k, l, t, n, cnt = 0;
    for(i = 1; i <= MAX; i *= 2){
        for(j = 1; j <= MAX; j*= 3){
            if(i*j > MAX) break;
            for(k = 1; k <= MAX; k *= 5){
                if(i*j*k > MAX) break;
                for(l = 1; l <= MAX; l *= 7){
                    if((t = i*j*k*l) <= MAX){
                        a[cnt++] = t;
                    }
                }
            }
        }
    }
    qsort(a, 5842, sizeof(long long), cmp);
    while (scanf("%lld", &n) != EOF && n){
        if(n % 10 == 1 && n % 100 != 11)
             printf("The %lldst humble number is %lld.\n", n, a[n-1]);
        else if(n % 10 == 2 && n % 100 != 12)
             printf("The %lldnd humble number is %lld.\n", n, a[n-1]);
        else if(n % 10 == 3 && n % 100 != 13)
             printf("The %lldrd humble number is %lld.\n", n, a[n-1]);
        else printf("The %lldth humble number is %lld.\n", n, a[n-1]);
    }
    return 0;
}

你可能感兴趣的:(ZOJ 1095 Humble Numbers)