蓝桥杯 ALGO-120 算法训练学做菜

问题描述
涛涛立志要做新好青年，他最近在学做菜。由于技术还很生疏，他只会用鸡蛋，西红柿，鸡丁，辣酱这四种原料来做菜，我们给这四种原料标上字母A,B,C,D。
涛涛现在会做的菜有五种：
1、西红柿炒鸡蛋原料：AABDD
2、酸辣鸡丁原料：ABCD
3、宫保鸡丁原料：CCD
4、水煮西红柿原料：BBB
5、怪味蛋原料：AD
这天早上，开开去早市给涛涛买了一些原料回来。由于事先没有什么计划，涛涛决定，对于现存的原料，每次尽量做菜单上靠前（即编号小）的菜。
现在请你写一个程序，判断一下开开和涛涛中午能吃到哪些菜。
输入格式
共4个整数a,b,c,d。分别表示开开买的A,B,C,D这4种原料的数量。每种原料不会超过30份。
输出格式
输出5行。其中第i行表示涛涛做的第i种菜的数目。
样例输入
3
1
2
4
样例输出
1
0
1
0
1

#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    if (a >= 2 && b >= 1 && d >= 2) {
        int minn1 = min(a / 2, b / 1);
        int minn = min(minn1, d / 2);
        a -= minn * 2;
        b -= minn * 1;
        d -= minn * 2;
        cout << minn << endl;
    }else{
        cout << 0 << endl;
    }
    if (a >= 1 && b >= 1 && d >= 1) {
        int minn1 = min(a / 1, b / 1);
        int minn2 = min(c / 1, d / 1);
        int minn = min(minn1, minn2);
        a -= minn * 1;
        b -= minn * 1;
        c -= minn * 1;
        d -= minn * 1;
        cout << minn << endl;
    }else{
        cout << 0 << endl;
    }
    if (c >= 2 && d >= 1) {
        int minn = min(c / 2, d / 1);
        c -= minn * 2;
        d -= minn * 1;
        cout << minn << endl;
    }else{
        cout << 0 << endl;
    }
    if (b >= 3) {
        int minn = b / 3;
        c -= minn * 3;
        cout << minn << endl;
    }else{
        cout << 0 << endl;
    }
    if(a >= 1 && d >= 1){
        int minn = min(a / 1, d / 1);
        cout << minn << endl;
    }else{
        cout << 0 << endl;
    }
    return 0;
}

#include <iostream>

#include <cmath>

using namespace std;

int main() {

int a, b, c, d;

cin >> a >> b >> c >> d;

if (a >= 2 && b >= 1 && d >= 2) {

int minn1 = min(a / 2, b / 1);

int minn = min(minn1, d / 2);

a -= minn * 2;

b -= minn * 1;

d -= minn * 2;

cout << minn << endl;

}else{

cout << 0 << endl;

}

if (a >= 1 && b >= 1 && d >= 1) {

int minn1 = min(a / 1, b / 1);

int minn2 = min(c / 1, d / 1);

int minn = min(minn1, minn2);

a -= minn * 1;

b -= minn * 1;

c -= minn * 1;

d -= minn * 1;

cout << minn << endl;

}else{

cout << 0 << endl;

}

if (c >= 2 && d >= 1) {

int minn = min(c / 2, d / 1);

c -= minn * 2;

d -= minn * 1;

cout << minn << endl;

}else{

cout << 0 << endl;

}

if (b >= 3) {

int minn = b / 3;

c -= minn * 3;

cout << minn << endl;

}else{

cout << 0 << endl;

}

if(a >= 1 && d >= 1){

int minn = min(a / 1, d / 1);

cout << minn << endl;

}else{

cout << 0 << endl;

}

return 0;

}

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