HDU100题简要题解（2060~2069）这⼗题感觉是100题内相对较为⿇烦的，好多数学题，有点搞我⼼态...歪⽐巴⼘HDU2060 SnookerProblem Descriptionbackground:Philip likes to play the QQ game of Snooker when he wants a relax, though he was just a little vegetable-bird. Maybe you hadn't played that game yet, no matter, I'll introduce the rule for you first.There are 21 object balls on board, including 15 red balls and 6 color balls: yellow, green, brown, blue, pink, black.The player should use a white main ball to make the object balls roll into the hole, the sum of the ball's fixed value he made in the hole is the player's score. The player should firstly made a red ball into the hole, after that he gains red-ball's value(1 points), then he gets the chance to make a color ball, then alternately. The color ball should be took out until all the red-ball are in the hole. In other word, if there are only color balls left on board, the player should hit the object balls in this order: yellow(2 point), green(3 point), brown(4 point), blue(5 point), pink(6 point), black(7 point), after the ball being hit into the hole, they are not get out of the hole, after no ball left on board, the game ends, the player who hasthe higher score wins the game. PS: red object balls never get out of the hole.I just illustrate the rules that maybe used, if you want to contact more details, visit afterthe contest.for example, if there are 12 red balls on board(if there are still red ball left on board, it can be sure that all the colorballs must be on board either). So suppose Philp can continuesly hit the ball into the hole, he can get the maximun score is12 * 1 (12 red-ball in one shoot) + 7 * 12(after hit a red ball, a black ball which was the most valuable ball should be the target) +2 +3 +4 +5 +6 + 7(when no red ball left, make all the color ball in hole).Now, your task is to judge whether Philip should make the decision to give up when telling you the condition on board(How many object balls still left not in the hole and the other player's score). If Philp still gets the chance to win, just print "Yes", otherwise print "No". (PS: if the max score he could get on board add his current score is equal to the opponent's current score, still output "Yes")InputThe first line contains a numble N indicating the total conditions. Then followed by N lines, each line is made of three integers: Ball_Left P_Score O_Score represeting the ball number left on board, Philp's current score, and the opponent's current score.All the input value are in 32 bit integer value range.OutputYou should caculate the max score left Philp can gain, and judge whether he has the possiblity to win.Sample Input212 1 11 30 39Sample OutputYesNo看到题⽬，脸上笑嘻嘻，⼼⾥......硬着头⽪看完了，发现是道⽔题：)分两种情况，加起来：1、⼤于六个球也就是有红球的情况，每打⼀个红球就紧接着打⼀个⿊球，打完红球再把其他颜⾊的打进去2、少于六个球也就是⽆红球的情况，假装剩下的都是相对较⼤的球#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int n, x, num1, num2;int main() {while (scanf("%d", &n) != EOF) {while (n--) {scanf("%d%d%d", &x, &num1, &num2);if (x > 6)num1 += (x - 6) * 8 + 27;else {int cnt = 7;while (x--) {num1 += cnt;cnt--;}}if (num1 >= num2) printf("Yes\n");else printf("No\n");}}return 0;}HDU2061 Treasure the new start, freshmen!Problem Descriptionbackground:A new semester comes , and the HDU also meets its 50th birthday. No matter what's your major, the only thing I want to tell youis:"Treasure the college life and seize the time." Most people thought that the college life should be colorful, less presure.But in actual, the college life is also busy and rough. If you want to master the knowledge learned from the book, a great deal of leisure time should be spend on individual study and practise, especially on the latter one. I think the every one of you should take the learning attitude just as you have in senior school."No pain, No Gain", HDU also has scholarship, who can win it? That's mainly rely on the GPA(grade-point average) of the student had got. Now, I gonna tell you the rule, and your task is to program to caculate the GPA.If there are K(K > 0) courses, the i-th course has the credit Ci, your score Si, then the result GPA isGPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)If there is a 0 <= Si < 60, The GPA is always not existed.InputThe first number N indicate that there are N test cases(N <= 50). In each case, there is a number K (the total courses number), then K lines followed, each line would obey the format: Course-Name (Length <= 30) , Credits(<= 10), Score(<= 100).Notice: There is no blank in the Course Name. All the Inputs are legalOutputOutput the GPA of each case as discribed above, if the GPA is not existed, ouput:"Sorry!", else just output the GPA value which is rounded to the 2 digits after the decimal point. There is a blank line between two test cases.Sample Input23Algorithm 3 97DataStruct 3 90softwareProject 4 852Database 4 59English 4 81Sample Output90.10Sorry!头⼤，看完发现⼜是⼀道⽔题，题意的精华在于其中的公式：GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)然后⽤这个算，并且⽤⼀个flag记录是否存在有不及格科⽬的情况#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int n, m;double a[101], b[101];int main() {while (scanf("%d", &n) != EOF) {while (n--) {scanf("%d", &m);double sum = 0;double num = 0;int flag = 0;for (int i = 1; i <= m; i++) {string s;cin >> s;scanf("%lf%lf", &a[i], &b[i]);if (b[i] < 60) flag = 1;sum += a[i] * b[i];num += a[i];}if (flag == 1) cout << "Sorry!" << endl;else {double ans = (sum + 0.0) / num;printf("%.2lf\n", ans);}if (n) cout << endl;}}return 0;}HDU2062 Subset sequenceProblem DescriptionConsider the aggregate An= { 1, 2, …, n }. For example, A1={1}, A3={1,2,3}. A subset sequence is defined as a array of a non-empty subset. Sort all the subset sequece of An in lexicography order. Your task is to find the m-th one.InputThe input contains several test cases. Each test case consists of two numbers n and m ( 0< n<= 20, 0< m<= the total number of the subset sequence of An ).OutputFor each test case, you should output the m-th subset sequence of An in one line.Sample Input1 12 12 22 32 43 10Sample Output111 222 12 3 1⼈傻了，康了题解才解决这题，卡这题上了⼀上午...这题我感觉算是这100题⾥思维含量最⾼的⼏个了，反正我是傻了（因为菜借这题终于明⽩了三元运算符，⾼中的时候看学弟⽤过，⼀脸懵逼也没想着去学，终于知道是个啥东西了（求gcd的时候还挺⽅便的，直接⼀⾏解决）题⽬⼤意：考虑⼀个集合 An = { 1, 2, ..., n}。
