HDU Problem - 1058 Humble Numbers 【dp】
来源:互联网 发布:php 微信扫码支付接口 编辑:程序博客网 时间:2024/05/29 09:28
Humble Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24415 Accepted Submission(s): 10702
Problem Description
A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers.
Write a program to find and print the nth element in this sequence
Write a program to find and print the nth element in this sequence
Input
The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.
Output
For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output.
Sample Input
1234111213212223100100058420
Sample Output
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.
Source
University of Ulm Local Contest 1996
Recommend
JGShining | We have carefully selected several similar problems for you: 1069 1421 1024 1081 1505
这道题主要是前面的打表部分,一开始想用一个4层循环解决,但是乘积会爆掉long long,然后百度知道用dp做。
#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "using namespace std;typedef long long LL;//typedef __int64 Int;typedef pair<int, int> paii;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double PI = acos(-1.0);const int MOD = 2000000000 + 7;const int MAXN = 10000 + 10;LL ans[MAXN];int p2 = 1, p3 = 1, p5 = 1, p7 = 1;LL min_f(LL x1, LL x2, LL x3, LL x4) {LL minn = min(min(x2, x1), min(x3, x4));//下面的式子不可以是if else 结构if (minn == x1) p2++;if (minn == x2) p3++;if (minn == x3) p5++;if (minn == x4) p7++;return minn;}void init() {ans[1] = 1LL;for (int i = 2; i <= 5842; i++) {ans[i] = min_f(ans[p2]*2LL, ans[p3]*3LL, ans[p5]*5LL, ans[p7]*7LL);}}int main() {init(); int n;while (scanf("%d", &n), n) {if (n%100 != 11 && n%10 == 1) printf("The %dst", n); else if (n % 100 != 12 && n % 10 == 2) printf("The %dnd", n); else if (n % 100 != 13 && n % 10 == 3) printf("The %drd", n); else printf("The %dth", n); printf(" humble number is %lld.\n", ans[n]);}return 0;}
0 0
- HDU Problem - 1058 Humble Numbers 【dp】
- hdu 1058 Humble Numbers (DP)
- hdu 1058 Humble Numbers (dp)
- hdu 1058 Humble Numbers(dp)
- hdu 1058 Humble Numbers(DP)
- 【DP】HDU-1058 Humble Numbers
- hdu 1058 Humble Numbers【dp】
- hdu 1058 Humble Numbers(dp)
- HDU 1058 Humble Numbers dp
- HDU 1058Humble Numbers(dp)
- HDU 1058 Humble Numbers DP
- HDU 1058 Humble Numbers 【DP】
- HDU 1058 Humble Numbers (DP)
- hdu 1058 Humble Numbers (DP)
- hdu 1058 Humble Numbers dp基础
- HDU 1058 Humble Numbers(DP,数)
- HDU 1058 Humble Numbers(dp+greedy)
- HDU 1058 Humble Numbers (dp+打表)
- (原创)分享一个自定义的tablayout组件
- 快速排序(C++实现)
- iconFont Access-Control-Allow-Origin
- 《算法》第二章——归并排序实现,快速归并,间接(索引)归并
- softmax回归
- HDU Problem - 1058 Humble Numbers 【dp】
- 如何学习嵌入式系统(基于ARM平台)
- 基于开源dxflib解析库的dxf文件显示
- Save results to different files when executing multi SQL statements in DB Query Analyzer 7.01
- wsacreateevent 事件使用
- Java程序执行过程的内存分析
- 定时向数组前添加不重复随机数
- tensorflow学习笔记(十九):分布式Tensorflow
- HTTP-Session&Cookie