UVa 654 - Ratio (枚举)

题意

这题目读得我蛋疼。

简单地说，就是先给一组数据，让我们从分母为1开始，逐步增加精度，如果发现某个数精度大于上一个精度，就输出。

思路

枚举分母暴力！

用round得出分子，然后把结果和上一次的比较。

因为可能出现可以约分的分数，我就用了gcd。。

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cctype>
   
   
   
   
#include <string>
   
   
   
   
#include <map>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define SZ(x) (int)x.size()
   
   
   
   
#define Lowbit(x) ((x) & (-x))
   
   
   
   
#define MP(a, b) make_pair(a, b)
   
   
   
   
#define MS(arr, num) memset(arr, num, sizeof(arr))
   
   
   
   
#define PB push_back
   
   
   
   
#define F first
   
   
   
   
#define S second
   
   
   
   
#define ROP freopen("input.txt", "r", stdin);
   
   
   
   
#define MID(a, b) (a + ((b - a) >> 1))
   
   
   
   
#define LC rt << 1, l, mid
   
   
   
   
#define RC rt << 1|1, mid + 1, r
   
   
   
   
#define LRT rt << 1
   
   
   
   
#define RRT rt << 1|1
   
   
   
   
#define BitCount(x) __builtin_popcount(x)
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const int MAXN = 150 + 10;
   
   
   
   
const int MOD = 1e9 + 7;
   
   
   
   
//const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
   
   
   
   

   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 // ROP;
   
   
   
   
 double win, lose;
   
   
   
   
 int i, j;
   
   
   
   
 bool first = true;
   
   
   
   
 while (~scanf("%lf%lf", &win, &lose))
   
   
   
   
 {
   
   
   
   
 if (!first) puts("");
   
   
   
   
 first = false;
   
   
   
   
 double cri = win / lose, last = INF;
   
   
   
   
 for (i = 1; i <= lose; i++)
   
   
   
   
 {
   
   
   
   
 double k = round(win * i / lose);
   
   
   
   
 if (__gcd((int)k, i) != 1) continue;
   
   
   
   
 if (fabs(cri - k / i) < last)
   
   
   
   
 {
   
   
   
   
 printf("%d/%d\n", (int)k, i);
   
   
   
   
 last = fabs(cri - k / i);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}