杭电ACM基础题目（1000 1001 1004 1005 1008 1012 1014）

1001、Sum Problem

Code:
1、注意：输出格式的要求，因为题目要求空一行，所以需要cout<<endl<<endl;
2、计算从1-n的和并输出

#include<iostream>
using namespace std;
int main() {
   
    int n,sum;
    while (cin >> n) {
   
        sum = 0;
        if (n == 0) return 0;
        for (int i = 1; i <= n; i++) {
   
            sum = sum + i;
        }
        cout << sum << endl<<endl;
    }

}

1004、 Let the Ballon Rise

Input
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) – the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.
A test case with N = 0 terminates the input and this test case is not to be processed.
Output
For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.
Sample Input

5
green
red
blue
red
red
3
pink
orange
pink
0

Sample Output

red
pink

Code:
1、二维数组输入字符串，字符串的长度有限定
2、strcmp()函数
3、比较数组中元素相同的个数

#include<iostream>
using namespace std;
int main() {
   
    int n;
    while (cin >> n) {
   
        if (n == 0) return 0;
        int num[1000] = {
   0};
        char color[1000][15];//借用二维数组输入字符串
        for (int i = 0; i < n; i++) {
   
            cin >> color[i];
        }
        //寻找color[]中个数最多的元素
        for (int i = 0; i < n; i++) {
   
            for (int j = i + 1; j < n; j++) {
   
                if (strcmp(color[i], color[j]) == 0) {
   //strcmp()字符串比较函数
                    num[i]++;
                }
            }
        }
        int max =0;
        for (int i = 1; i < n; i++) {
   
            if (num[i] > max)
                max = i;
        }
        cout << color[max] << endl;
    }

}

1005、 Number Sequence

Problem Description
A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.
Sample Input

1 1 3
1 2 10
0 0 0

Sample Output

2
5

Code:
方法一：使用递归的方式，但会出现内存超限的问题

#include<iostream>
using namespace std;
int f(int a, int b, int n) {
   
    int result;
    if (n == 1||n==2) {
   
        return 1;
    }
    else {
   
        result=(a * f(a,b,n - 1) + b *