HDU1058 Humble Numbers
2013-01-30 16:23
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题目:Humble Numbers
humble number从1为"始祖",剩下的所有数,其实都是在此基础上乘以2,3,5,7演化出来的,代码主要语句:f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);
humble number从1为"始祖",剩下的所有数,其实都是在此基础上乘以2,3,5,7演化出来的,代码主要语句:f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);
#include <iostream> #include <stdio.h> using namespace std; int f[5843],n; int i,j,k,l; int min(int a,int b,int c,int d) { int min=a; if(b<min) min=b; if(c<min) min=c; if(d<min) min=d; if(a==min) i++; if(b==min) j++; if(c==min) k++; if(d==min) l++; return min; } int main() { i=j=k=l=1; f[1]=1; for(int t=2;t<=5842;t++) { f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]); } while(scanf("%d",&n)&&n!=0) { if(n%10==1&&n%100!=11) printf("The %dst humble number is %d.\n",n,f ); else if(n%10==2&&n%100!=12) printf("The %dnd humble number is %d.\n",n,f ); else if(n%10==3&&n%100!=13) printf("The %drd humble number is %d.\n",n,f ); else printf("The %dth humble number is %d.\n",n,f ); } return 1; }
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