hdu1058(找第n个数,其因子只含2,3,5,7)

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题目意思:找第n个数,其因子只含2,3,5,7.

#include<stdio.h>
int main()
{int i,i1,i2,i3,i4,a[5850],min,min1;a[1]=1;i1=i2=i3=i4=1;for(i=2;i<=5842;i++)//打表算出拆分的最简因子只包含2,3,5,7四个数{min=a[i1]*2<a[i2]*3?a[i1]*2:a[i2]*3;//这些数从小到大排列min1=a[i3]*5<a[i4]*7?a[i3]*5:a[i4]*7;min=min<min1?min:min1;a[i]=min;if(min==a[i1]*2) i1++;if(min==a[i2]*3) i2++;if(min==a[i3]*5) i3++;if(min==a[i4]*7) i4++;}int n;while(scanf("%d",&n)==1&&n){if(n%10==1&&n%100!=11)printf("The %dst humble number is %d.\n",n,a[n]);else if(n%10==2&&n%100!=12)printf("The %dnd humble number is %d.\n",n,a[n]);else if(n%10==3&&n%100!=13)printf("The %drd humble number is %d.\n",n,a[n]);elseprintf("The %dth humble number is %d.\n",n,a[n]);}return 0;
}

 

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