踩气球,气球上的分值为1-100间的数,参赛者的初始分数为1,每踩一个气球就把气球的分数乘以自己的分数,得到比分,最后选手会声明自己的比分,但是有的人可能计算错误,所以最后的分数可能不正确,判断一下两个分数是否能产生….
刚接触acm的时候就看到这个题了..当时是什么思路也没有,现在又看还是不会,我总是想着把,所有的分解情况都列出来再进行比较,但是貌似不可行.今天忍不住偷偷看了别人的代码(貌似我已经几个月没有看过别人的代码了),顿时就震精了,为什么我想不到呢? 😥
#include<iostream>
using namespace std;
int aa,bb;
void dfs(int a,int b,int n)
{
if(b==1)
{
if(a==1) aa=1;
bb=1;
}
if( n > 100 || aa==1 && bb==1 )
return;
if(b%n==0) dfs(a,b/n,n+1);//除或不除,他就在那里,一直递归....
if(a%n==0) dfs(a/n,b,n+1);
dfs(a,b,n+1);
}
int main(void)
{
int a,b,t;
while( cin >> a >> b )
{
if(a < b)
{t=a;a=b;b=t;}
aa=bb=0;
dfs(a,b,2);
if(aa==1)
cout << a;
else if(bb==1)
cout << b;
else
cout << a;
cout << endl;
}
return 0;
}
忍不住…姐，你需要脑子再灵活点哈～
@ZXY: 你转街的时候还上来看这些阿…YM ➡
这个代码的确很牛X，不过在你点开无数blog，发现都是同一段代码时，会觉得很悲催……
@laoda: 在acmdiy群里面发现你了，你就是那个“早起一水”的【zzuli】laoda吧。嘿嘿。。
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不会呀。。。 👿 👿
你什么时候在实验室？
很好奇一个case：
input：4 8
output：8
这算不算一个bug？
求教