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  • HDU_1058 Humble Numbers(DP)

      动态规划题,F[n] 可由2, 3, 5, 7与F[n-1]的乘积得到,取这四个数的的最小值,就是当前F[n]的值,即F[n] = min{F[n-1]*2, F[n-1]*3, F[n-1]*5, F[n-1]*7}。

      前6个数的递归过程:

           n1 = 1;

      n2 = min{1*2, 1*3,1*5,1*7} = 2;

      n3 = min{2*2, 1*3, 1*5, 1*7} = 3;

      n4 = min{2*2, 2*3, 1*5, 1*7} = 4;

      n5 = min{3*2, 2*3, 1*5, 1*7} = 5;

      n6 = min{3*2, 2*3, 2*5, 1*7} = 6;

      代码如下:

    #include <stdio.h>
    int min(int x, int y)
    {
    return x < y ? x : y;
    }
    int main()
    {
    int i, a, b, c, d, n;
    int num[5850];
    num[
    1] = 1;
    a
    = b = c = d = 1;
    for(i = 2; i <= 5842; i++)
    {
    num[i]
    = min(num[a]*2, min(num[b]*3, min(num[c]*5, num[d]*7)));
    if(num[i] == num[a]*2) a++;
    if(num[i] == num[b]*3) b++;
    if(num[i] == num[c]*5) c++;
    if(num[i] == num[d]*7)   d++;
    }
    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", num[n]);
    }
    return 0;
    }
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  • 原文地址:https://www.cnblogs.com/vongang/p/2126837.html
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