题目六：2055.An easy problem
Problem Description
we define f(A) = 1, f(a) = -1, f(B) = 2, f(b) = -2, ... f(Z) = 26, f(z) = -26;
Give you a letter x and a number y , you should output the result of y+f(x).
Input
On the first line, contains a number T.then T lines follow, each line is a case.each case contains a letter and a number.
Output
for each case, you should the result of y+f(x) on a line.
#include <iostream> #include <vector> using namespace std; int main() { int m, n; char c; vector<int> num(26); for (int i = 0; i < 26; i++) { num[i] = i + 1; } while (cin >> m) { while (m--) { cin >> c >> n; if (c >= 'a' && c <= 'z') { n = n - num[c - 'a']; } else if (c >= 'A' && c <= 'Z') { n = n + num[c - 'A']; } cout << n << endl; } } return 0; }
题目七：2010.水仙花数
Problem Description
春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花数，他是这样定义的：
“水仙花数”是指一个三位数，它的各位数字的立方和等于其本身，比如：153=1^3+5^3+3^3。
现在要求输出所有在m和n范围内的水仙花数。
Input
输入数据有多组，每组占一行，包括两个整数m和n（100<=m<=n<=999）。
Output
对于每个测试实例，要求输出所有在给定范围内的水仙花数，就是说，输出的水仙花数必须大于等于m,并且小于等于n，如果有多个，则要求从小到大排列在一行内输出，之间用一个空格隔开;
如果给定的范围内不存在水仙花数，则输出no;
每个测试实例的输出占一行。
#include <iostream> #include <vector> #include <string> using namespace std; class NUM { public: NUM(int M, int N) : m_min(M), m_max(N) {} vector<int> find() const { vector<int> result; for (int n = m_min; n <= m_max; ++n) { if (NN(n)) { result.push_back(n); } } return result; } private: int m_min, m_max; bool NN(int number) const { int sum = 0; string digits =to_string(number); for (char digit : digits) { sum += pow(digit - '0', 3); } return sum == number; } }; int main() { int m, n; while (cin >> m >> n) { if (100 <= m && m <= n && n <= 999) { NUM finder(m, n); vector<int> NUMs = finder.find(); if (!NUMs.empty()) { for (size_t i = 0; i < NUMs.size(); ++i) { cout << NUMs[i]; if (i < NUMs.size() - 1) cout << " "; } cout << endl; } else { cout << "no" <<endl; } } } return 0; }
题目八：2057.A+B Again
Problem Description
There must be many A + B problems in our HDOJ , now a new one is coming.
Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too.
Easy ? AC it !
Input
The input contains several test cases, please process to the end of the file.
Each case consists of two hexadecimal integers A and B in a line seperated by a blank.
The length of A and B is less than 15.
Output
For each test case,print the sum of A and B in hexadecimal in one line.
#include<iostream> #include <cmath> #include <cstdlib> #include <algorithm> using namespace std; int main(void){ __int64 a,b,sum; while(cin>>hex>>a&&cin>>hex>>b){ sum=a+b; if(sum<0){ sum=-sum; cout<<hex<<uppercase<<"-"<<sum<<endl; }else{ cout<<hex<<uppercase<<sum<<endl; } } }
题目九：2011.多项式求和
Problem Description
多项式的描述如下：
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...
现在请你求出该多项式的前n项的和。
Input
输入数据由2行组成，首先是一个正整数m（m<100），表示测试实例的个数，第二行包含m个正整数，对于每一个整数(不妨设为n,n<1000），求该多项式的前n项的和。
Output
对于每个测试实例n，要求输出多项式前n项的和。每个测试实例的输出占一行，结果保留2位小数。
#include <iostream> #include <iomanip> using namespace std; double Sum(int n) { double sum = 0.0; for (int i = 1; i <= n; ++i) { if (i % 2 == 1) { sum += 1.0 / i; // 对于奇数项，进行加法操作 } else { sum -= 1.0 / i; // 对于偶数项，进行减法操作 } } return sum; } int main() { int m; cin >> m; for (int i = 0; i < m; ++i) { int n; cin >> n; cout << fixed << setprecision(2) << Sum(n) << endl; } return 0; }
题目十：2012.素数判定
Problem Description
对于表达式n^2+n+41，当n在（x,y）范围内取整数值时（包括x,y）(-39<=x<y<=50)，判定该表达式的值是否都为素数。
Input
输入数据有多组，每组占一行，由两个整数x，y组成，当x=0,y=0时，表示输入结束，该行不做处理。
Output
对于每个给定范围内的取值，如果表达式的值都为素数，则输出"OK",否则请输出“Sorry”,每组输出占一行。
#include <iostream> using namespace std; bool Prime(int num) { if (num <= 1) return false; if (num <= 3) return true; if (num % 2 == 0 || num % 3 == 0) return false; for (int i = 5; i * i <= num; i += 6) { if (num % i == 0 || num % (i + 2) == 0) return false; } return true; } int main() { int x, y; while (cin >> x >> y) { if (x == 0 && y == 0) break; bool all= true; for (int n = x; n <= y; ++n) { int value = n * n + n + 41; if (!Prime(value)) { all= false; break; } } if (all) cout << "OK" << endl; else cout << "Sorry" << endl; } return 0; }
