codeforces Round #184 Div.2 - B Continued Fractions

一开始直接用递归求后一个分数式的值，每求一步约分一次，过了小数据，大数据没过……后来发现是因为中间两个long long相乘溢出了……

看了别人的代码才明白，只要思维稍微转化一下，这题就能变得很简单……

 

 1 #include <cstring>

 2 #include <cstdlib>

 3 #include <cstdio>

 4 #include <algorithm>

 5 

 6 using namespace std;

 7 

 8 #define LL long long int

 9 

10 const int MAXN = 100;

11 

12 int n;

13 LL p, q, a[MAXN];

14 

15 int main()

16 {

17     //freopen("in.txt", "r", stdin );

18     //freopen("out.txt", "w", stdout );

19     while ( ~scanf( "%I64d%I64d", &p, &q ) )

20     {

21         int i;

22         scanf( "%d", &n );

23         for ( i = 0; i < n; ++i )

24             scanf( "%I64d", &a[i] );

25 

26         for ( i = 0; i < n; ++i )

27         {

28             if ( a[i] > p / q || ( i + 1 < n && a[i] * q == p ) ) break;

29             p -= q * a[i];

30             swap( p, q );

31         }

32 

33         if ( i == n && !q ) puts("YES");

34         else puts("NO");

35     }

36     return 0;

37 }