解题思路：一道三分搜索的水题，可直接使用模板！
解题代码：
View Code
1 // File Name: Hurry Up 2 // Author: sheng 3 // Created Time: 2013年05月12日 星期日 20时36分10秒 4 5 #include <stdio.h> 6 #include <string.h> 7 #include <math.h> 8 #include <iostream> 9 using namespace std; 10 const double EPS = 1e-10; 11 12 int x1, yy1, x2, yy2, vr, vt; //y1为c++库函数的变量所以用yy1 13 14 double calc(double x) 15 { 16 return (sqrt( (x1 - x) * (x1 - x) + yy1 * yy1)/ vr + sqrt( (x2 - x) * (x2 - x) + yy2 * yy2)/vt); 17 } 18 19 double solve () 20 { 21 double xx1 = (double)x1; 22 double xx2 = (double)x2; 23 if (xx1 > xx2) 24 { 25 double temp = xx1; 26 xx1 = xx2; 27 xx2 = temp; 28 } 29 // printf ("%lf %lf\n", xx1, xx2); 30 double left = xx1, right = xx2; 31 double mid, midmid; 32 double mid_value, midmid_value; 33 while (left + EPS < right) 34 { 35 mid = (left + right)/2; 36 midmid = (mid + right)/2; 37 mid_value = calc(mid); 38 midmid_value = calc(midmid); 39 // cout << mid_value <<" " << midmid_value << endl; 40 if (mid_value < midmid_value) 41 right = midmid; 42 else left = mid; 43 } 44 return calc(left); 45 } 46 47 int main () 48 { 49 int cas; 50 double ans1, ans2; 51 scanf ("%d", &cas); 52 while (cas--) 53 { 54 scanf ("%d%d%d%d%d%d", &x1, &yy1, &x2, &yy2, &vr, &vt); 55 ans1 = sqrt ( (x1 - x2) * (x1 - x2) + (yy1 - yy2) * (yy1 - yy2))/vr; 56 ans2 = solve(); 57 // printf ("%.2lf %.2lf\n", ans1, ans2); 58 printf ("%.2lf\n", ans1 > ans2 ? ans2 : ans1); 59 } 60 return 0; 61 }
Jack’s sequence
题目链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1177
解题思路：此题我们只需找出我们序列中间最后一个“（”变成“）”之后依旧能与之前匹配的位置然后直接输出就可以了，具体方法看代码吧！
解题代码：
View Code
1 // File Name: Jack’s sequence 1177.cpp 2 // Author: sheng 3 // Created Time: 2013年05月12日 星期日 21时58分06秒 4 5 #include <stdio.h> 6 #include <iostream> 7 #include <string.h> 8 using namespace std; 9 10 const int max_n = 10010; 11 char bg[max_n]; 12 int sign[max_n]; 13 14 int main () 15 { 16 int n, tag; 17 int cnt; 18 int cun; 19 scanf ("%d", &n); 20 getchar(); 21 while (n--) 22 { 23 cnt = cun = 0; 24 scanf ("%s", bg); 25 memset (sign, 0, sizeof (sign)); 26 int len = strlen (bg); 27 for (int i = 0; i < len; i ++) 28 { 29 if (bg[i] == '(') 30 { 31 cnt ++; 32 tag = 0; 33 } 34 else 35 { 36 cnt --; 37 if (cnt > 0 && !tag) 38 sign[cun++] = tag = i - 1; 39 } 40 } 41 tag = 0; 42 cnt = 0; 43 for (int i = cun - 1; i >= 0; i --) 44 { 45 if ( sign[i]) 46 { 47 tag = sign[i]; 48 break; 49 } 50 } 51 if (!tag) 52 { 53 cout << "No solution\n"; 54 continue; 55 } 56 for (int i = 0; i < tag; i ++) 57 { 58 cout << bg[i]; 59 if (bg[i] == '(') 60 cnt ++; 61 else cnt --; 62 } 63 cnt --; 64 cout << ")"; 65 for (int i = 0; i < (len - cnt - tag)/2; i ++) 66 cout << "("; 67 for (int i = 0; i < (len - tag + cnt)/2; i ++) 68 cout << ")"; 69 cout << endl; 70 } 71 return 0; 72 }