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;
}