Humble Numbers 1095
摘要:
View Code #include <iostream>using namespace std;int a[5844];int min(int b,int c,int d,int e){ int t; t=b<c?b:c; t=t<d?t:d; t=t<e?t:e; return t;}int main(){ int i=0; a[i]=1; int x=0,y=0,z=0,m=0; while(i<=5843) { a[++i]=min(a[x]*2,a[y]*3,a[z]*5,a[m]*7); if(... 阅读全文
posted @ 2012-04-10 22:37 不悔梦归处 阅读(170) 评论(0) 推荐(0) 编辑