Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 14933 Accepted
Submission(s): 6494
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#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#define MAX 5843using
namespace std;int find_min(int
a , int
b , int
c , int
d){ int
temp = a < b ? a : b ; int
index = c < d ? c : d ; return
(temp < index ? temp : index) ;}int
main(){ int
i,j,n,ta,tb,tc,td,id1,id2,id3,id4; int
dp[MAX]; dp[0]=1; id1=id2=id3=id4=0; for(j=1; j<MAX; j++) { dp[j]=find_min(ta=dp[id1]*2,tb=dp[id2]*3,tc=dp[id3]*5,td=dp[id4]*7); if(dp[j]==ta) id1++; if(dp[j]==tb) id2++; if(dp[j]==tc) id3++; if(dp[j]==td) id4++; } while(cin>>n&&n) { if(n%10==1&&n%100!=11) { printf("The %dst humble number is %d.\n",n,dp[n-1]); continue; } if(n%10==2&&n%100!=12) { printf("The %dnd humble number is %d.\n",n,dp[n-1]); continue; } if(n%100!=13&&n%10==3) { printf("The %drd humble number is %d.\n",n,dp[n-1]); continue; } printf("The %dth humble number is %d.\n",n,dp[n-1]); } return
0;} |
原文：http://www.cnblogs.com/lxm940130740/p/3582693.html