C++程序设计实践一下（题目来自杭州电子科技大学ACM）

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