计算糖果

#include<iostream>
using namespace std;

int main()
{   int a, b, c, d;
    int A, B, C;
	while(cin>>a>>b>>c>>d)
	{
		A=(a+c)/2;
		B=(b+d)/2;
		C=(d-b)/2;
		if(A-B==a && B-C==b && A+B==c && B+C==d)
		{
			cout<<A<<" "<<B<<" "<<C<<endl;
		}
		else
		{
			cout<<"No"<<endl;
		}
	}
    return 0;
}

//由于涉及到除法，得到的结果A,B,C有可能是截断后的数据，不妨逆推回去验算一遍，假如与原方程相符合，则正确

// 要求ABC都为整数
// 注意多了一个约束条件，即四个式子解3个未知数，所以还要确定满不满足A+B==n3

import java.util.Scanner;
public class Main {	
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int n1 = in.nextInt();
		int n2 = in.nextInt();
		int n3 = in.nextInt();
		int n4 = in.nextInt();
		int A = (n1+n3)/2;
		int B = (n2+n4)/2;
		int C = B-n2;
		if((n1+n3)%2==0&&(n2+n4)%2==0&&(A+B==n3)){
			System.out.println(A+" "+B+" "+C);
		}else{
			System.out.println("No");
		}
	}
}
 

#include<iostream>
using namespace std;
int main()
{
    int x1, x2, x3, x4;
    cin >> x1 >> x2 >> x3 >> x4;
    if(x1+x2+x4 != x3)//判断是否无解-自相矛盾
        cout << "No" << endl;
    else
    {
        int B = (x2+x4) / 2;
        if(2*B != (x2+x4))//判断是否为非整数解
        {
            cout << "No" << endl;
        }
        else
        {
            int A = B + x1;
            int C = B - x2;
            cout << A << " " << B  << " "<< C << endl;
        }
    }
}

	

增广矩阵要有唯一解，要求r(A|b) = r(A)，可以得到一个式子，不等于0时则没有唯一解。满足则可以
根据阶梯矩阵求解A B C。
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int M = scanner.nextInt();
        int N = scanner.nextInt();
        int P = scanner.nextInt();
        int Q = scanner.nextInt();
        if (M - P + N + Q != 0) {
            System.out.println("No");
        }else {
            int A = P - (N + Q) / 2;
            int B = (N + Q) / 2;
            int C = (Q - N) / 2;
            System.out.println(A + " " + B + " " + C);
        }
    }
}

python四行解法

這道題很簡單，不過要注意算出來的值不能爲負數，不能是小數。

i1, i2, i3, i4 = map(int, input().split())
a = (i1 + i3) // 2 if (i1 + i3) % 2 == 0 and (i1 + i3) >= 0 else None
b = (i2 + i4) // 2 if (i2 + i4) % 2 == 0 and (i2 + i4) >= 0 else None
print(" ".join(map(str, [a, b, i4 - b])) if isinstance(a, int) and isinstance(b, int) else "No")

#include <iostream>
#include <vector>
using namespace std;
int main()
{
    vector<int> vec(4);
    int a, b, c;
    for(int i = 0; i < 4; ++i)
    {
        cin >> vec[i];
    }
    a = (vec[0] + vec[2])/2;
    b = (vec[1] + vec[3])/2;
    c = (vec[3] - vec[1])/2;
    if( ((a-b) == vec[0]) && ((b-c) == vec[1]) && ((a+b) == vec[2])
          && ((b+c) == vec[3]) )
        cout << a << " " << b << " " << c << endl;
    else 
        cout << "No" << endl;
    return 0;
}

#include<stdio.h>
int main()
{
    int A,B,C,a,b,c,d;
    scanf("%d%d%d%d",&a,&b,&c,&d);
    A = (a+c)/2;
    B = (a-c)/(-2);
    C = (b-d)/(-2);
    if(B == (b+d)/2)
        printf("%d %d %d",A,B,C);
    else
        printf("No");
}

python 2 line solution

a, b, c, d = list(map(int,input().strip().split()))
print((a+c)//2,(b+d)//2,(d-b)//2) if c-a == b+d else print('No')

#include<stdio.h>
int main()
{
 int n1,n2,n3,n4,A1,B1,C1;
 float A,B,C;
 scanf("%d %d %d %d",&n1,&n2,&n3,&n4);
 A=(n1+n3)/2.0;
 B=(n2+n4)/2.0;
 C=n4-B;
 A1=(int)A;
 B1=(int)B;
 C1=(int)C;
 if((A1==A)&&(B1==B)&&(C1==C)&&(A1>=0)&&(B1>=0)&&(C1>=0)) 
 printf("%d %d %d",A1,B1,C1);
 else 
 printf("No");
 return 0;
}

<?php
//输入1 -2 3 4
//输出 2 1 3
$arr = explode(' ',trim(fgets(STDIN)));
$a = ($arr[0]+$arr[2])/2; //A-B+A+B = 2A =1+3=4   A=2
$b = ($arr[1]+$arr[3])/2; //B-C+B+C = 2B =-2+4=2  B=1
$c = $arr[3]-$b;  // B+C=4  C=4-B
//糖果数必须是大于等于0的整数
if(!is_int($a*$b*$c)||$a<0||$b<0||$c<0)
    echo 'No';
else
    echo "$a $b $c";

def isOK(x):
    if x < 0: return False
    if int(x) == x:  #是否是小数
        return True

line = raw_input()
data = line.strip().split(" ")
data = map(int,data)

d0 = data[0]
d1 = data[1]
d2 = data[2]
d3 = data[3]

#求解A、B、C
A  = 0.5 * (d2 + d0)
B1 = 0.5 * (d2 - d0)
C  = 0.5 * (d3 - d1)
B2 = 0.5 * (d3 + d1)  #B1、B2为B的两种算法，必须一致才能满足要求

if isOK(A) and isOK(B1) and isOK(C) and (B1 == B2):
    A = int(A)
    B1= int(B1)
    C = int(C)
    print str(A) +" " + str(B1) + " " + str(C)
else:
    print "No"

	

import java.util.Scanner;
public class Main{
    public static void main(String args[]){
        int result1,result2,result3,result4;
        Scanner sc = new Scanner(System.in);
        result1 = sc.nextInt();
        result2 = sc.nextInt();
        result3 = sc.nextInt();
        result4 = sc.nextInt();
        int A,B,C;
        if((result1+result3)%2==0){
            A=(result1+result3)/2;
            B=result3-A;
            C=result4-B;
            if(result1==A-B && result2==B-C && result3==A+B && result4==B+C){
                System.out.print(A+" "+B+" "+C);
            }else{
                System.out.print("No");
            }
        }else{
            System.out.print("No");
        }
    } 
}

//根据代数式特点，直接用消元法可以得到结果
#include<iostream>
using namespace std;
int main(){
    int a,b,c,d;
    cin>>a>>b>>c>>d;
    int A=(int)(a+c)/2;
    int B=(int)(b+d)/2;
    int C=(int)(d-b)/2;
    if(A-B==a&&B-C==b&&A+B==c&&B+C==d){
        cout<<A<<" "<<B<<" "<<C<<endl;
    }
    else
        cout<<"No"<<endl;
    return 0;
}

#include <iostream>

using namespace std;

int main()
{     int a,b,c,d;     int A,B,C;     while(cin>>a>>b>>c>>d)     {         A = (a+c)/2;         B = (b+d)/2;         C = (d-b)/2;         if(A-B==a && B-C==b && A+B==c && B+C==d)             cout<<A<<" "<<B<<" "<<C<<endl;         else             cout<<"No"<<endl;     }     return 0;
}

#include<stdio.h>
int main(){
    int i,j,k,flag=0,a,b,c,d,res1,res2,res3;
    scanf("%d%d%d%d",&a,&b,&c,&d);
    for(i=-30;i<=30;i++)
        for(j=-30;j<=30;j++)
            for(k=-30;k<=30;k++)
                if(i-j==a&&j-k==b&&i+j==c&&j+k==d){
                    flag=1;
                    res1=i,res2=j,res3=k;
                }
    flag?printf("%d %d %d\n",res1,res2,res3):printf("No\n");
}

import java.util.Scanner;

public class StringUtil {

	//计算糖果
	public static void main(String[ ] args){
		
		Scanner in = new Scanner(System.in);
		while(in.hasNext()){
			int a = in.nextInt();
			int b = in.nextInt();
			int c = in.nextInt();
			int d = in.nextInt();
			int x = (c+a)/2;
			int y = (c-a)/2;
			int z = (c-a-2*b)/2;
			if(x-y == a && x+y == c && y-z == b && y+z == d && x*2 == a+c && y*2 == c-a && z*2 == c-a-2*b){
				System.out.println(x + " " + y + " " + z);
			}
			else{
				System.out.println("No");
			}
		}
	}
}