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