题目大意：给出n条线段，把每条线段变成原来线段的一条子线段，使得改变后的所有线段等长且不相交，输出最大长度

解题思路：这题要用long double确保精度，精度要求很高，接下来就是将这个long double的数转化为两个整数相除的结果了，有两种方法，学习到了

#include
   
     #include
    
      #include
     
       using namespace std; #define maxn 100010 #define esp 1e-11 struct Line{ int L, R; }line[maxn]; int n; bool ok(long double len) { long double cur = 0.0, t; for(int i = 0; i < n; i++) { t = line[i].L * 1.0; cur = max(t, cur); if(cur + len - line[i].R > esp) return false; cur += len; } return true; } bool cmp(const Line a, const Line b) { if(a.L == b.L) return a.R < b.R; return a.L < b.L; } int main() { while(scanf("%d", &n) != EOF) { for(int i = 0; i < n; i++) scanf("%d%d", &line[i].L, &line[i].R); sort(line, line + n, cmp); long double L = 0.0, R = line[n - 1].R + esp; while(R - L > esp) { long double mid = (L + R) / 2; if(ok(mid)) L = mid; else R = mid; } /* int i; for(i = 1; i <= n + 1; i++) { int p = (int)(L * i + 0.5); long double t = p * 1.0 / i; if(t - L < 2 * esp && L - t < esp * 2) break; } printf("%d/%d\n", (int)(L * i + 0.5), i); */ long double Min = 0x3f3f3f3f; int p, q; for(int i = 1; i <= maxn; i++) { int j = floor(L * i); if(fabs((long double)(j) / i - L) < Min) { p = j; q = i; Min = fabs((long double)(j) / i - L); } j = ceil(L * i); if(fabs((long double)(j) / i - L) < Min) { p = j; q = i; Min = fabs((long double)(j) / i - L); } } printf("%d/%d\n", p, q); } return 0; }