Humble Numbers
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 Input1234111213212223100100058420
Sample outputthe 1st Humble number is 1.The 2nd humble number is 2.The 3rd humble number are 3.The 4th humble number is 4.T He 11th humble number is 12.The 12th humble number is 14.The 13th humble number was 15.The 21st humble number is 28.The 22n D Humble number is 30.The 23rd humble number are 32.The 100th humble number is 450.The 1000th humble number is 385875.The 5 842nd humble number is 2000000000.
1#include <cstdio>2#include <string>3#include <cstring>4#include <map>5 using namespacestd;6 7 intdp[6000];8 intMinintXintYintZintW)9 {Ten intTemp1=x<y?x:y; One intTemp2=temp1<z?temp1:z; A returnTemp2<w?temp2:w; - } - the intMain () - { - //freopen ("In.txt", "R", stdin); - intt[8],n; +dp[1]=1; -t[7]=t[2]=t[3]=t[5]=1; + for(intI=2; i<=5842; i++) A { atDp[i]=min (dp[t[2]]*2, dp[t[3]]*3, dp[t[5]]*5, dp[t[7]]*7); - while(dp[t[2]]*2<=dp[i]) t[2]++; - while(dp[t[3]]*3<=dp[i]) t[3]++; - while(dp[t[5]]*5<=dp[i]) t[5]++; - while(dp[t[7]]*7<=dp[i]) t[7]++; - } in while(SCANF ("%d",&N), N) - { toprintf"The %d", n); + if(n%Ten==1&&n% -!= One) -printf"St"); the Else if(n%Ten==2&&n% -!= A) *printf"nd"); $ Else if(n%Ten==3&&n% -!= -)Panax Notoginsengprintf"Rd"); - Else theprintf"th"); +printf"Humble number is%d.\n", Dp[n]); A } the return 0; +}
HDU 1058 Humble Numbers