如下图所示的三角形,有三个边s1,s2,s3,边s1有四个圆圈⭕️1,2,3,4,边s2有四个圆圈⭕️4,5,6,7,边s3有四个圆圈⭕️7,8,9,1 把1-9这9个数字,分别填写到下图所示的9个圆圈圆圈里, 使每条边上的4个圆圈⭕️的和相等(9个圆圈里的数字不能重复) 例如:s1=[1=>9,2=>8,3=>1,4=>3]=9+8+1+3=21,s2=[4=>3,5=>5,6=>7,7=>6]=3+5+7+6=21,s3=[7=>6,8=>2,9=>4, 1=>9]=6+2+4+9=21
如下图所示的三角形,有三个边s1,s2,s3,边s1有四个圆圈⭕️1,2,3,4,边s2有四个圆圈⭕️4,5,6,7,边s3有四个圆圈⭕️7,8,9,1 把1-9这9个数字,分别填写到下图所示的9个圆圈圆圈里, 使每条边上的4个圆圈⭕️的和相等(9个圆圈里的数字不能重复) 例如:s1=[1=>9,2=>8,3=>1,4=>3]=9+8+1+3=21,s2=[4=>3,5=>5,6=>7,7=>6]=3+5+7+6=21,s3=[7=>6,8=>2,9=>4, 1=>9]=6+2+4+9=21
输入内容格式:3,4,8,5,2,6,7,1,9其中3,4, 8,5 是s1 1-4填写的数字,数字之和205,2,6,7 是s2 4-7填写的数字,数字之和207,1,9,3 是s4 7-1填写的数字,数字之和20
当输入的数字满足三边之和相等的时候,输出yes当输入的数字不满足三边之和相等的时候,输出no
3,4,8,5,2,6,7,1,9
yes
3,4,8,5,2,6,9,7,1
no
var nums = []int{1, 2, 3, 4, 5, 6, 7, 8, 9} //允许填写的数字var result [][]int //合法结果集func permute(nums []int) [][]int {
//write code here
}
#include <stdio.h> #include <stdlib.h> int main(){ int num[9]; char c[9]; int i; for(i = 0;i<9;i++){ scanf("%d,%s",&num[i],c[i]); } if(num[0]+num[1]+num[2]+num[3] == num[3]+num[4]+num[5]+num[6] &&num[3]+num[4]+num[5]+num[6] == num[6]+num[7]+num[8]+num[0]) { printf("yes"); }else{ printf("no"); } return 0; }
import java.util.*; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String i = scanner.next(); String[] str = i.split(","); ArrayList<Integer> list = new ArrayList<>(); for (String s : str) { list.add(Integer.parseInt(s)); } int n1 = list.get(0) + list.get(1) + list.get(2) + list.get(3); int n2 = list.get(3) + list.get(4) + list.get(5) + list.get(6); int n3 = list.get(6) + list.get(7) + list.get(8) + list.get(0); if (n1 == n2 && n2 == n3) { System.out.println("yes"); } else { System.out.println("no"); } } }
def sum(str): a=str[0]+str[1]+str[2]+str[3] b = str[3] + str[4] + str[5] + str[6] c = str[6] + str[7] + str[8] + str[0] if a==b and a==c and b==c: print('yes') else: print('no') #例子:3,4,8,5,2,6,7,1,9 yes 3,4,8,5,2,6,9,7,1 no n=input() n=n.split(',') str=[] for i in n: str.append(int(i)) sum(str)
#大致思路,三边相等的话,求出平均数肯定等于其中一边 def sum(*args): one=0 c=0 #s1的和 for i in range(len(args)): one=one+args[i] if i == 3: #s1的边长 c=one one=one+args[i] elif i ==6: #s2的边长 one = one + args[i] elif i==8: #s3边长 one=one+args[0] if one / 3 == c: print('yes') else: print('no') #a=(int(i) for i in input('please input number:')) # print(a) a=(3,4,8,5,2,6,7,1,9) sum(*a)
print('please input the 9 numbers:')L=input()print('the input is ', L)X=list(L)firstArr = X[0:4]secArr = X[3:7]thirArr = X[6:9] + list(X[0])firstSum = 0secSum = 0thirdSum = 0for i in range(0, 4):firstSum += int(firstArr[i])for i in range(0, 4):secSum += int(secArr[i])for i in range(0, 4):thirdSum += int(thirArr[i])if firstSum == secSum and secSum ==thirdSum:print('Yes')else:print('No')