标签：acm   hdu   动态规划   dp   

题意  所有只能被2，3，5，7这4个素数整除的数称为Humble Number  输入n  输出第n个Humble Number

1是第一个humble number  对于一个Humble Number  a  有2*a,3*a,5*a,7*a都是Humble Number  可以以1为基数  依次展开即可得到一定范围内的Humble Number 用i,j,k,l分别记录 2，3，5，7分别乘到了第几个Humble Number  当前在计算第cnt个Humble Number  那么有  hum[cnt] = min ( hum[i] * 2, hum[j] * 3, hum[k] * 5, hum[l] * 7)   然后对应min的i或j或k或l就加1  当cnt到达了n  结果就出来了

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 5843;
int hum[N], cnt, n;
int main()
{
    int i = 1, j = 1, k = 1, l = hum[1] = 1;
    for (cnt = 2; cnt < N; ++cnt)
    {
        hum[cnt] = min ( min(hum[i] * 2, hum[j] * 3), min (hum[k] * 5, hum[l] * 7));
        if (hum[cnt] == hum[i] * 2) ++i;
        if (hum[cnt] == hum[j] * 3) ++j;
        if (hum[cnt] == hum[k] * 5) ++k;
        if (hum[cnt] == hum[l] * 7) ++l;
    }
    while (scanf ("%d", &n), n)
    {
        printf ("The %d", n);
        if (n % 100 != 11 && n % 10 == 1) printf ("st ");
        else if (n % 100 != 12 && n % 10 == 2) printf ("nd ");
        else if (n % 100 != 13 && n % 10 == 3) printf ("rd ");
        else printf ("th ");
        printf ("humble number is %d.\n", hum[n]);
    }
    return 0;
}

Humble Numbers

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.

HDU 1058 Humble Numbers（DP,数）,布布扣,bubuko.com

HDU 1058 Humble Numbers（DP,数）

标签：acm   hdu   动态规划   dp   

原文地址：http://blog.csdn.net/iooden/article/details/38511119