Hdoj-1058-humble Numbers "dp&"

Source: Internet
Author: User

Humble NumbersTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 20715 Accepted Submission (s): 9045


Problem Descriptiona number whose only prime factors be 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, ten,, +, (+), A, ten, A, ten, ten, ... shows the first humble numbers.

Write a program to find and print the nth element in this sequence

Inputthe input consists of one or more test cases. Each test case consists the one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for N.

Outputfor each test case, print one line saying "The nth humble number is number." Depending on the value of n, the correct suffix "st", "nd", "ts", or "th" for the ordinal number nth have to is used like I T 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 are 3.The 4th humble number is 4.The 11th HUMBL E number is 12.The 12th humble number are 14.The 13th humble number is 15.The 21st humble number is 28.The 22nd humble numb ER is 30.The 23rd humble number is 32.The 100th humble number are 450.The 1000th humble number is 385875.The 5842nd humble Number is 2000000000.

Sourceuniversity of Ulm Local Contest 1996
recommendjgshining | We have carefully selected several similar problems for you:1024 1025 1081 1160 2870
#include <stdio.h>int dp[5900];int min (int a,int b,int c,int d) {int T1,t2;t1=a>b?b:a;t2=c>d?d:c;return t1 >T2?T2:T1;} void f () {int f1,f2,f3,f4;f1=f2=f3=f4=1;dp[1]=1;for (int i=2;i<=5842;++i) {dp[i]=min (dp[f1]*2,dp[f2]*3,dp[f3]*5, DP[F4]*7); if (dp[i]==dp[f1]*2) f1++;if (dp[i]==dp[f2]*3) f2++;if (dp[i]==dp[f3]*5) f3++;if (dp[i]==dp[f4]*7) f4++;}} int main () {f (); int N;while (scanf ("%d", &n), N) {printf ("The%d", n); if (n%10==1&&n%100!=11) printf ("St"); else if (n%10==2&&n%100!=12) printf ("nd"), else if (n%10==3&&n%100!=13) printf ("rd"), else printf ("th") ;p rintf ("Humble number is%d.\n", Dp[n]);} return 0;}


Good question, just start not to find ideas, look at other people's code just understand, really water ah

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Hdoj-1058-humble Numbers "dp&"

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