AtCoder Beginner Contest 088 CTakahashi's Information 【思维枚举】

C - Takahashi’s Information

Time limit : 2sec / Memory limit : 256MB Score: 300 points

Problem Statement

We have a 3×3 grid. A number ci,j is written in the square (i,j), where (i,j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a1,a2,a3,b1,b2,b3 whose values are fixed, and the number written in the square (i,j) is equal to ai+bj. Determine if he is correct.

Constraints

•ci,j (1≤i≤3,1≤j≤3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format: c1,1 c1,2 c1,3 c2,1 c2,2 c2,3 c3,1 c3,2 c3,3

Output

If Takahashi’s statement is correct, print ‘Yes’; otherwise, print ‘No’.

Sample Input 1

1 0 1 2 1 2 1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as: a1=0,a2=1,a3=0,b1=1,b2=0,b3=1.

Sample Input 2

2 2 2 2 1 2 2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8 0 8 8 0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6 2 9 7 0 7 7

Sample Output 4

No

题意 ： 给你一个3*3的矩阵c，然后问你是否存在 6个数 : a1,a2,a3,b1,b2,b3使得 c[i][j] = ai + bj

分析：我们可以只枚举a1，根据第一列和第一行的矩阵的值，反推出 a2,a3,b1,b2,b3,然后在判断 c[2][3],c[2][2],c[3][3],c[3][2]是否符合，范围呢，你可以选择大些，我只选择了 [-1000,1000]

参考代码

#include<bits/stdc++.h> using namespace std; int a[4][4]; int main(){ for(int i = 1; i<= 3;i++) { for(int j = 1;j <= 3;j ++) { cin>>a[i][j]; } } for(int k = -1000;k <= 1000;k++) { int a2,a3,b1,b2,b3; int a1 = k; b1 = a[1][1] - k; b2 = a[1][2] - k; b3 = a[1][3] - k; a2 = a[2][1] - b1; a3 = a[3][1] - b1; bool flg = false; if(a[2][2] != a2 + b2) flg = true; if(a[2][3] != a2 + b3) flg = true; if(a[3][2] != a3 + b2) flg = true; if(a[3][3] != a3 + b3) flg = true; if(!flg) { cout<<"Yes"<<endl;return 0; } } cout<<"No"<<endl; return 0; } 如有错误或遗漏，请私聊下UP，thx