题解 | #买蛋糕#

F. 买蛋糕

知识点：分支、循环

比较入门，没什么好讲的。

由于 的范围不大，如果还没学到循环，可以手敲 个分支。

标程

C++

循环版

#include <bits/stdc++.h>

using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; ++i)
    {
        int a, b, c;
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    cout << ans << '\n';
}

没学过循环版

#include <bits/stdc++.h>

using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, a, b, c, ans = 0;
    cin >> n;
    if (n >= 1)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 2)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 3)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 4)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 5)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 6)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 7)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 8)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 9)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 10)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 11)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 12)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 13)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 14)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 15)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 16)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 17)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 18)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 19)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    if (n >= 20)
    {
        cin >> a >> b >> c;
        if (a + b + c >= 100)
            ++ans;
    }
    cout << ans << '\n';
}

Java

import java.util.*;
import java.io.*;

public class Main {
    static Kattio io = new Kattio();

    public static void main(String[] args) {
        int n = io.nextInt();
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int a = io.nextInt();
            int b = io.nextInt();
            int c = io.nextInt();
            if (a + b + c >= 100) {
                ans++;
            }
        }
        io.println(ans);

        io.close();
    }
}

class Kattio extends PrintWriter {
    private BufferedReader r;
    private StringTokenizer st;
    // 标准 IO
    public Kattio() { this(System.in, System.out); }
    public Kattio(InputStream i, OutputStream o) {
        super(o);
        r = new BufferedReader(new InputStreamReader(i));
    }
    // 在没有其他输入时返回 null
    public String next() {
        try {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(r.readLine());
            return st.nextToken();
        } catch (Exception e) {}
        return null;
    }
    public int nextInt() { return Integer.parseInt(next()); }
    public double nextDouble() { return Double.parseDouble(next()); }
    public long nextLong() { return Long.parseLong(next()); }
}

Python

n = int(input())
ans = 0

for i in range(n):
    a, b, c = map(int, input().split())
    if a + b + c >= 100:
        ans += 1

print(ans)