基础题集题解 -2021.1.15

A - A + B Problem HDU - 1000

水题

#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
   
   
	ios::sync_with_stdio(0);
	int a, b;
	while (cin >> a >> b)
		cout << a + b << endl;
	return 0;
}

B - Sum Problem HDU - 1001

水题

#include<map>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
   
   
	ios::sync_with_stdio(0);
	int a, b, sum = 0;
	while (cin >> b) {
   
   
		sum = 0;
		for (int i = 1; i <= b; i++)
			sum += i;
		cout << sum << endl << endl;
	}
	return 0;
}

C - Let the Balloon Rise HDU - 1004

使用map记录每个字符串的出现次数即可

#include<map>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
   
   
	ios::sync_with_stdio(0);
	map<string,int> m;
	int num;
	while (cin >> num){
   
   
		if (num == 0)
			break;
		string max;
		int cnt_max = 0;
		m.clear();
		while (num--) {
   
   
			string s;
			cin >> s;
			m[s]++;
			if (cnt_max < m[s]) {
   
   
				cnt_max = m[s];
				max = s;
			}
		}
		cout << max << endl;
	}
	return 0;
}

D - Number Sequence HDU - 1005

矩阵快速幂运算，需先计算出初始矩阵，还有一种打表找规律的解法，可交题ac，但题目本质上是递推公式

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2;
const int mod = 7;
struct matrix {
   
   
	long long m[maxn][maxn];
	matrix() {
   
   
		memset(m, 0, sizeof(m));
	}
};
matrix multi(matrix a, matrix b) {
   
   
	matrix res;
	for (int i = 0; i < maxn; i++)
		for (int j = 0; j < maxn; j++)
			for (int k = 0; k < maxn; k++)
				res.m[i][j] = res.m[i][j] + a.m[i][k] * b.m[k][j] % mod;
	return res;
}
matrix fastm(matrix a, int n) {
   
   
	matrix res;
	for (int i = 0; i < maxn; i++)
		res.m[i][i] = 1;
	while (n) {
   
   
		if (n & 1)
			res = multi(res, a);
		a = multi(a, a);
		n >>= 1;
	}
	return res;
}
int main()
{
   
   
	matrix mar;
	long long x, y, n;
	while (cin >> x >> y >> n) {
   
   
		if (x == 0 && y == 0 && n == 0)
			break;
		mar.m[0][0] = x;
		mar.m[0][1] = y;
		mar.m[1][0] = 1;
		mar.m[1][1] = 0;
		if (n <= 2)
			cout << "1" << endl;
		else {
   
   
			matrix res = fastm(mar, n - 2);
			cout << (res.m[0][0] + res.m[0][1]) % 7 << endl;
		}
	}
	return 0;
}

E - Elevator HDU - 1008

水题

#include<stdio.h>
int main(int argc, const char* argv[])
{
   
   
	int sum, num = 0,floor,floor1 = 0;;
	while (scanf("%d", &num) != EOF)
	{
   
   
		if (num == 0)
			break;
		sum = 0;
		floor = 0;
		for (int i = 0; i < num; i++)
		{
   
   
			scanf("%d", &floor1);
			if (floor > floor1)
				sum