题目 Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence Eva must c…
题目 On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen.Now given a string that you are supposed to type, and the string that you actually type out,…
题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer taking all the characters in S2 from S1. Your task is simply to calculate S1 – S2 for any given strings. However, it might not be that simple to do it fast. Input…
题目 Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 10^4]. The first one who bets on a unique number wins. For example, if there ar…
The task of this problem is simple: insert a sequence of distinct positive integers into a hash table first. Then try to find another sequence of integer keys from the table and output the average search time (the number of comparisons made to find w…
题目 编程团体赛的规则为:每个参赛队由若⼲队员组成:所有队员独⽴⽐赛:参赛队的成绩为所有队员的成绩和:成绩最⾼的队获胜.现给定所有队员的⽐赛成绩,请你编写程序找出冠军队. 输⼊格式: 输⼊第⼀⾏给出⼀个正整数N(<=10000),即所有参赛队员总数.随后N⾏,每⾏给出⼀位队员的成绩,格式为:"队伍编号-队员编号 成绩",其中"队伍编号"为1到1000的正整数,"队员编号"为1到10的正整数,"成绩"为0到100的整数.…
PAT (Advanced Level) Practice 1031 Hello World for U (20 分) 凌宸1642 题目描述: Given any string of N (≥5) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as: h d e l l r lowo That is, the charact…
PAT (Advanced Level) Practice 1023 Have Fun with Numbers (20 分) 凌宸1642 题目描述: Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be…
简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; int n; ]; int num1,num2;…
因为会溢出,因此判断条件需要转化.变成b>c-a #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; long long a,b,c; int main() { int T; scanf("%d",&T);…